Using a standard Normal distribution, what is the cut off z-score for the top 24%? a)-1.08 Ob) -0.71 Oc) 1.08 d) 0.71

Answer:

135

Step-by-step explanation:

A straight line is 180° and the total sum of interior angles of a triangle is 180° too. So the missing angle (inside the triangle) here = 180 - (60+75) = 180 - 135 = 45°

But x = 180 - missing angle (as the total of x plus the missing angle should be 180)

So x = 180 - 45 = 135°

The answer is option (a) 0.2312.

The standard deviation of the sampling distribution of the sample mean is determined using the formula below:

σ(ȳ) = σ/√nwhere:σ(ȳ) is the standard deviation of the sample means,

σ is the population standard deviation,n is the sample size.

The mean of the sample means is the same as the population mean.

The Central Limit Theorem (CLT) ensures that the distribution of sample means is usually normal.

Using the formula above:σ(ȳ) = σ/√nσ(ȳ)

= 10/√6σ(ȳ) = 4.08

The standard deviation of the sampling distribution of the sample means is 4.08, whereas the population mean is 200. To find the probability that their mean IQ is greater than 197, we need to compute the z-score.

z = (x - μ) / σ(ȳ)where:

x = 197μ = 200σ(ȳ) = 4.08z = (197 - 200) / 4.08z = -0.736

Round to two decimal places:-0.74

The area to the right of -0.74 under the standard normal distribution curve is:0.7704

We need to subtract this from 1 to obtain the probability that their mean IQ is greater than 197.

P(Z > -0.74) = 1 - 0.7704 = 0.2296

Round to four decimal places.0.2296Therefore, the answer is option (a) 0.2312.

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Q1 is located approximately 0.6745 standard deviations below the mean and Q3 is located approximately 0.6745 standard deviations above the mean.

a. The distribution of X, the age that children learn to walk, is normally distributed.

X - N(13, 1.5)

b. The distribution of x, the sample mean age of the 15 randomly selected people, is also normally distributed.

x - N(13, 1.5/sqrt(15))

c. To find the probability that one randomly selected person learned to walk between 12.5 and 14 months old, we can standardize the values using the z-score formula and then look up the probabilities in the standard normal distribution table.

P(12.5 ≤ X ≤ 14) = P((12.5 - 13) / 1.5 ≤ Z ≤ (14 - 13) / 1.5)

Standardizing the values:

P(-0.3333 ≤ Z ≤ 0.6667)

Looking up the probabilities in the standard normal distribution table, we find the corresponding values:

P(-0.3333 ≤ Z ≤ 0.6667) ≈ P(Z ≤ 0.6667) - P(Z ≤ -0.3333)

Using the table or a calculator, we find:

P(-0.3333 ≤ Z ≤ 0.6667) ≈ 0.7461 - 0.3694 ≈ 0.3767

Therefore, the probability that one randomly selected person learned to walk between 12.5 and 14 months old is approximately 0.3767.

d. For the 15 people, to find the probability that the average age they learned to walk is between 12.5 and 14 months old, we can use the same method as in part c, but with the distribution of the sample mean.

P(12.5 ≤ x ≤ 14) = P((12.5 - 13) / (1.5/sqrt(15)) ≤ Z ≤ (14 - 13) / (1.5/sqrt(15)))

Standardizing the values:

P(-1.2247 ≤ Z ≤ 1.2247)

Looking up the probabilities in the standard normal distribution table, we find the corresponding values:

P(-1.2247 ≤ Z ≤ 1.2247) ≈ P(Z ≤ 1.2247) - P(Z ≤ -1.2247)

Using the table or a calculator, we find:

P(-1.2247 ≤ Z ≤ 1.2247) ≈ 0.8904 - 0.1096 ≈ 0.7808

Therefore, the probability that the average age the 15 people learned to walk is between 12.5 and 14 months old is approximately 0.7808.

e. Yes, the assumption that the distribution is normal is necessary for part d). The reason is that the probability calculations for the sample mean rely on the Central Limit Theorem, which states that the distribution of the sample mean approaches a normal distribution as the sample size increases. In this case, with a sample size of 15, the assumption of normality is necessary for the probability calculation.

f. To find the interquartile range (IQR) for the average first-time walking age for groups of 15 people, we need to calculate the first quartile (Q1) and the third quartile (Q3) for the distribution of the sample mean.

Using the properties of the normal distribution, we know that Q1 is located approximately 0.6745 standard deviations below the mean and Q3 is located approximately 0.6745 standard deviations above the mean.

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a. The probability of a pregnancy lasting 309 days or longer is approximately 0.0035.

b. Babies born on or before 240 days are considered premature.

a. The probability of a pregnancy lasting 309 days or longer can be found by calculating the z-score and using the standard normal distribution table. First, we calculate the z-score using the formula:

z = (x - μ) / σ

where x is the desired value (309 days), μ is the mean (269 days), and σ is the standard deviation (15 days). Substituting the values, we get:

z = (309 - 269) / 15 = 40 / 15 = 2.67

Next, we look up the z-score of 2.67 in the standard normal distribution table and find the corresponding probability. The table shows that the area to the left of the z-score of 2.67 is approximately 0.9965. However, we are interested in the probability of the pregnancy lasting 309 days or longer, so we subtract this value from 1:

P(X ≥ 309) = 1 - 0.9965 = 0.0035

Therefore, the probability that a pregnancy will last 309 days or longer is approximately 0.0035.

b. To find the length that separates premature babies from those who are not premature, we need to determine the value that corresponds to the lowest 3% in the distribution. This is equivalent to finding the z-score that has an area of 0.03 to its left. We look up this z-score in the standard normal distribution table and find it to be approximately -1.88.

To find the corresponding length, we use the z-score formula:

z = (x - μ) / σ

Rearranging the formula to solve for x:

x = μ + z * σ

Substituting the values, we have:

x = 269 + (-1.88) * 15 = 269 - 28.2 = 240.8

Rounding to the nearest integer, we conclude that babies born on or before 240 days are considered premature.

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The first question is about the relationship between opinion on gun ownership and race-ethnicity. 67% of White respondents, 28% of Black respondents, and 64% of Hispanic respondents shared the view that widespread gun ownership protects law abiding citizens from crime.

The opinion on gun ownership is dependent on race-ethnicity. Roughly 20% of undergraduates at a university are vegetarian or vegan. We need to find the probability that, among a random sample of 3 undergraduates, at least one is vegetarian or vegan.

P(at least one is vegetarian or vegan) = 1 - P(none of them is vegetarian or vegan) P(none of them is vegetarian or vegan)

= (0.8)^3

= 0.512

P(at least one is vegetarian or vegan) = 1 - 0.512

= 0.488

The probability that, among a random sample of 3 undergraduates, at least one is vegetarian or vegan is 0.488.

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We can say that with 90% confidence we estimate that the true population mean number of personal computers per household is between 1.3858 and 1.5742.

A simple random sample of 195 households was taken, and the sample mean number of personal computers was 1.48. The population standard deviation is a 0.8. The number of computers is being calculated here

(a)The formula for constructing the confidence interval is:

CI= x ± z* (σ/√n)

Here, the sample mean is given as x = 1.48

Population standard deviation σ = 0.8

Sample size n = 195

The 90% confidence interval means that alpha (α) = 1 - 0.9 = 0.1 on either side.The z-value for alpha/2 = 0.05 is 1.645.Then substituting the values in the formula,

CI = 1.48 ± 1.645 * (0.8/√195)

CI = 1.48 ± 0.112

CI = (1.3858, 1.5742)

Thus, a 90% confidence interval for the mean number of personal computers is 1.3858 << 1.5742.

:Therefore, we can say that with 90% confidence we estimate that the true population mean number of personal computers per household is between 1.3858 and 1.5742.

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The domain of f(x) is all real numbers except for -7. The domain of g(x) is all real numbers.

The domain of a function is the set of all real numbers that can be input into the function without causing the function to be undefined.

In the case of f(x), the denominator of the fraction is equal to 0 when x = -7. Therefore, x cannot be equal to -7. This means that the domain of f(x) is all real numbers except for -7.

In the case of g(x), there are no real numbers that make the function undefined. Therefore, the domain of g(x) is all real numbers.

Here are the domains of f(x) and g(x) in interval notation:

f(x) : (-∞, -7) ∪ (-7, ∞)

g(x) : (-∞, ∞)

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The correct answer is Option d

To solve this problem, we need to calculate the probability of selecting all seven students as undergraduate students from the random sample.

Assuming that the selection of students is done without replacement (once a student is chosen, they cannot be chosen again), the probability of selecting an undergraduate student on the first pick is 1/2 since undergraduate and graduate students are equally represented.

After the first pick, there will be six remaining students, and the probability of selecting another undergraduate student from this reduced pool is also 1/2.

This process continues for the remaining picks until we have chosen all seven students.

To calculate the probability of all seven students being undergraduate students, we multiply the probabilities of each individual pick:

(1/2) * (1/2) * (1/2) * (1/2) * (1/2) * (1/2) * (1/2) = (1/2)^7 = 1/128 ≈ 0.0078

Therefore, the correct answer is option d. The probability that all seven students chosen are undergraduate students is approximately 0.0078.

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The function h(x) has two nonremovable discontinuities at x = 3 and x = -6. To find and classify the discontinuities of the function h(x) = (x^2 - 3x) / (x^2 + 3x - 18):

We need to determine if there are any points where the function is not defined or exhibits a jump, removable, or nonremovable discontinuity.

Step 1: Find the points where the function is not defined.

The function h(x) will be undefined when the denominator of the fraction becomes zero. So, we solve the equation x^2 + 3x - 18 = 0 to find the values of x that make the denominator zero.

Factoring the quadratic equation, we have (x - 3)(x + 6) = 0.

Therefore, the function is not defined at x = 3 and x = -6.

Step 2: Classify the discontinuities.

(a) Removable Discontinuities: To determine if a discontinuity is removable, we need to check if the limit of the function exists at that point and if it can be redefined or filled in to make the function continuous.

At x = 3 and x = -6, we evaluate the limit of h(x) as x approaches these points. If the limits exist and are finite, the discontinuities are removable.

(b) Nonremovable Discontinuities: If the limits do not exist or are infinite at a point, the discontinuity is nonremovable.

Step 3: Evaluate the limits at the discontinuity points.

(a) At x = 3:

To evaluate the limit as x approaches 3, we substitute x = 3 into the function h(x).

h(3) = (3^2 - 33) / (3^2 + 33 - 18) = 0 / 0.

Since the denominator is also zero, we have an indeterminate form, and further analysis is needed to determine if it is a removable or nonremovable discontinuity.

Taking the derivative of h(x), we get h'(x) = (-6x + 3) / (x^2 + 3x - 18).

Evaluating h'(x) at x = 3, we have h'(3) = 3 / 0, which is undefined.

Since the derivative is also undefined, the discontinuity at x = 3 is nonremovable.

(b) At x = -6:

Substituting x = -6 into the function h(x), we get h(-6) = (-6^2 - 3*(-6)) / (-6^2 + 3*(-6) - 18) = 0 / 0.

Similar to the previous case, we have an indeterminate form, and we need to examine the derivative of h(x) to determine the nature of the discontinuity.

Differentiating h(x), we obtain h'(x) = (6x - 3) / (x^2 + 3x - 18).

Evaluating h'(x) at x = -6, we have h'(-6) = -3 / 0, which is undefined.

Since the derivative is also undefined, the discontinuity at x = -6 is nonremovable.

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The probability that Z falls between 1.05 and 2.13 is 0.1307.

We know that a standard normal distribution has a mean of 0 and a standard deviation of 1. We want to find the probability that the random variable Z falls between 1.05 and 2.13.

To solve this problem, we need to find the area under the standard normal curve between the Z-scores of 1.05 and 2.13. We can use a standard normal table or calculator to find this probability.

Using a standard normal table, we can find the probability of Z being less than 2.13 and subtract the probability of Z being less than 1.05.

The value for Z = 2.13 can be looked up in the standard normal table and we find that the corresponding probability is 0.9838.

The value for Z = 1.05 can also be looked up and we find that the corresponding probability is 0.8531.

Therefore, P(1.05 ≤Z≤2.13) = P(Z ≤ 2.13) - P(Z ≤ 1.05) = 0.9838 - 0.8531 = 0.1307.

Therefore, the probability that Z falls between 1.05 and 2.13 is 0.1307.

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The value of x is 25° and the values of the angles is 52°

A circle is a special kind of ellipse in which the eccentricity is zero and the two foci are coincident.

In circle geometry, there is a theorem that states that; angle formed in the same segment are equal.

Therefore we can say that;

3x -23 = 2x +2

collect like terms

3x -2x = 23 +2

x = 25

Therefore the value of is 25 and each angle in the segment is calculated as

3x -23 = 3( 25 ) -23

75 -23

= 52°

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The given bell-shaped distribution has a mean of 250.6 and a standard deviation of 62.3. (All units are 1000 cells/uL.)Using the empirical rule, the approximate percentage of women with platelet counts within 1 standard deviation of the mean, or between 188.3 and 312.9 are 68%.

The approximate percentage of women with platelet counts within 1 standard deviation of the mean, or between 188.3 and 312.9 are 68%.Explanation:To calculate the percentage of women with platelet counts within 1 standard deviation of the mean, we will use the empirical rule. This rule is based on the normal distribution of data, according to which 68% of data falls within one standard deviation from the mean.The given bell-shaped distribution has a mean of 250.6 and a standard deviation of 62.3.

Therefore, 1 standard deviation from the mean can be calculated as follows:Lower limit = mean - standard deviation = 250.6 - 62.3 = 188.3Upper limit = mean + standard deviation = 250.6 + 62.3 = 312.9Thus, approximately 68% of women in this group have platelet counts within 1 standard deviation of the mean, or between 188.3 and 312.9 .b. The approximate percentage of women with platelet counts between 126.0 and 375.2 are 95%.Explanation:To calculate the percentage of women with platelet counts between 126.0 and 375.2, we will use the empirical rule. This rule is based on the normal distribution of data, according to which:68% of data falls within one standard deviation from the mean.95% of data falls within two standard deviations from the mean.99.7% of data falls within three standard deviations from the mean.

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The average daily rate of a hotel in Canada in August 2018 was $193.85. Assume that the average daily rate follows a normal distribution with a standard deviation of $29.80.

Using the formula for z-score, we can find the probabilities of the given events. P(X < 170). the population mean$\sigma$ is the population standard deviation $z$ is the z-score x is the given score By substituting the given values, we get:$z = \frac{170 - 193.85}{29.80}$$

z = -0.80$

Now, we have to find the probability P(X < 170). Since the given distribution is a normal distribution, we can find the probability using the standard normal distribution table. Using the table, we get: P(Z < -0.80) = 0.2119 Therefore, P(X < 170) = 0.2119 ii) P(X > 220)Using the formula for z-score, we get:$z = \frac{220 - 193.85}{29.80}

$z_1 = \frac{145 - 193.85}{29.80}$$z_1

= -1.63$$z_2

= \frac{190 - 193.85}{29.80}$$z_2

= -0.13$ Now, we have to find the probability P(145 < X < 190). Using the standard normal distribution table, we get:P(-1.63 < Z < -0.13) = 0.4236Therefore, P(145 < X < 190) = 0.4236 Hence, the probabilities of the given events are:

i) P(X < 170) = 0.211

ii) P(X > 220) = 0.1907

iii) P(145 < X < 190) = 0.4236

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The observations given are;7.3 7.0 6.5 7.5 7.6 6.3 6.9 7.6 6.4 6.9Mean and standard deviation of the data can be calculated by the following formulas; Mean,  X = (ΣX)/nStandard Deviation,  s = [Σ(X-X )²/(n-1)]Now, substitute the values and calculate as follows

Mean, X = (7.3+7+6.5+7.5+7.6+6.3+6.9+7.6+6.4+6.9)/

10 = 6.99 (rounded to two decimal places)Standard Deviation,

s = [((7.3-6.99)² + (7-6.99)² + (6.5-6.99)² + (7.5-6.99)² + (7.6-6.99)² + (6.3-6.99)² + (6.9-6.99)² + (7.6-6.99)² + (6.4-6.99)² + (6.9-6.99)²)/(10-1)]^(1/2) = 0.5496 (rounded to four decimal places)Therefore, mean is 6.99 and standard deviation is 0.5496. (b)Since the sample size is small (n < 30) and population standard deviation is unknown, we will use t-distribution for finding confidence interval. 99% upper one-sided confidence bound for the population mean will be;Upper one-sided confidence

bound = X + tα,df,s/√n

Where, X = 6.99 (sample mean)tα,df,

s = t

0.01,9,0.5496 = 2.8214 (obtained from t-distribution table for

α = 0.01,

df = n

-1 = 9)

s = 0.5496 (standard deviation)

n = 10 (sample size)∴ Upper one-sided confidence

bound = 6.99 + 2.8214*0.5496/

√10 = 7.4385 (rounded to three decimal places)Therefore, 99% upper one-sided confidence bound for the population mean is 7.439.

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The profit is maximized when quantity reaches 200 and average cost function is C(x) = 131/x + 7.4.

Given that cost and revenue are given by C(x) = 131 + 7.4x and R(x) = 8x - 0.01x², we have to find the average cost function.

Average cost function is given as; AC(x) = C(x)/x

The total cost is given by;C(x) = 131 + 7.4x

The total revenue is given by;

R(x) = 8x - 0.01x²

The quantity is given by x;

Average cost function is given by;

AC(x) = C(x)/x= (131 + 7.4x) / x= 131/x + 7.4

Let's verify this using calculus;

C(x) = 131 + 7.4xR(x) = 8x - 0.01x²

The profit is given by;

P(x) = R(x) - C(x)

P(x) = (8x - 0.01x²) - (131 + 7.4x)

P(x) = -0.01x² + 0.6x - 131

The marginal revenue (MR) is the derivative of R(x);

MR = dR(x) / dx

MR = 8 - 0.02x

The marginal cost (MC) is the derivative of C(x);

MC = dC(x) / dx

MC = 7.4

The profit is maximized when MR = MC;

8 - 0.02x = 7.4x = 200

The quantity is 200.

The average cost function is;

AC(x) = C(x)/x= (131 + 7.4x) / x= 131/x + 7.4

Therefore, the average cost function is C(x) = 131/x + 7.4.

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The following are the solutions to the given problem: The domain of f is R, that is, all real numbers.

It is denoted as D(f) = (-∞, ∞).

The reason is that f(x) is defined for all x, i.e. there is no restriction on x. 2. The critical values of f are (-2.094, 7.351, 1.743).

To obtain these values, we need to first compute f' (x), which is -3x² - 9x + 12.

Then, we set this equation equal to 0 and solve for x. After we obtain x, we check the value of f'' (x) to ensure that we have obtained critical points. If f'' (x) < 0, we have a local maximum, and if f'' (x) > 0, we have a local minimum.

If f'' (x) = 0, we have a point of inflection.

If we have multiple values of x, we need to check the sign of f'' (x) to identify the type of point. 3. The possible points of inflection of f (x-values only) are (-1.53, 60.31) and (0.423, 51.33).

To obtain these values, we need to first compute f'' (x), which is -6x - 9. Then, we set this equation equal to 0 and solve for x. After we obtain x, we substitute it into f(x) to obtain the corresponding y-value.

There are no vertical asymptotes for f(x). 5. The horizontal asymptote for f(x) is y = -∞. To determine this, we need to evaluate the limit of f(x) as x approaches infinity and negative infinity. As x approaches infinity, f(x) approaches negative infinity. As x approaches negative infinity, f(x) approaches positive infinity.

The number line analysis is as follows:

The critical points are (-2.094, 83.21), (1.743, 44.33), and (7.351, -45.68). The possible points of inflection are (-1.53, 60.31) and (0.423, 51.33).

The given function is f(x)=−x 3 −4.5x 2 +12x+52.

The domain of f is R, and there are no vertical asymptotes for f(x). The horizontal asymptote for f(x) is y = -∞.

The critical values of f are (-2.094, 7.351, 1.743).

The possible points of inflection of f (x-values only) are (-1.53, 60.31) and (0.423, 51.33).

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MedianThe formula to calculate the median of the given data is:Median = [(n+1)/2]th. The given data for sales and salesman is as follows:No of Sales: 0-4 5-9 10-14 15-19 20-24 25-29.

Number of Salesman: 1 14 23 21 15 6MeanThe formula to calculate the mean of the given data is:Mean (X) = (ΣX)/n where, ΣX is the sum of all the values in the given data and n is the total number of values.Mean = (0x1 + 5x14 + 10x23 + 15x21 + 20x15 + 25x6)/(1+14+23+21+15+6)= 371/80

Mean = 4.64 approx.MedianThe formula to calculate the median of the given data is:

Median = [(n+1)/2]th observation when n is odd

Median = ([n/2]th observation + [(n/2)+1]th observation)/2 when n is even.To calculate the median, we need to arrange the given data in ascending order:No of Sales: 0-4 5-9 10-14 15-19 20-24 25-29 Number of Salesman: 1 14 23 21 15 6Arranging in ascending order: No of Sales: 0-4 5-9 10-14 15-19 20-24 25-29 Number of Salesman: 1 14 15 21 23 6Median = ([n/2]th observation + [(n/2)+1]th observation)/2

when n is even= ([80/2]th observation + [(80/2)+1]th observation)/2

= (40th observation + 41st observation)/2

= (21 + 23)/2

Median = 22

Mode The mode is the value that appears most frequently in the given data. The value 10-14 appears most frequently in the given data. Therefore, the mode is 10-14.VarianceThe formula to calculate the variance of the given data is:σ2 = Σ(x - μ)2 / N where,σ2 is the variance,x is the value in the data set,μ is the mean of the data set,N is the total number of values.

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(a) The slope of the regression line is D. 4.372. (b) The correlation coefficient is A. 0.8398. (c) The alternate hypothesis is D. \( \mathrm{H}_{1}: \rho \neq 0 \). (d) The test statistic is C. 4.794. (e) The degrees of freedom are A. 11. (f) At the 5% significance level, it can be concluded that there is evidence to suggest the correlation coefficient is C. not zero.

(a) The slope of the regression line represents the change in the dependent variable (y) for every one unit increase in the independent variable (x). In this case, the slope is calculated as the coefficient of the independent variable from the regression analysis. From the given options, the slope is 4.372, indicating that for every one unit increase in the statistics of practice games (x), the statistics of official games (y) are expected to increase by 4.372 units. Therefore, the correct answer is D. 4.372.

(b) The correlation coefficient measures the strength and direction of the linear relationship between two variables. It ranges from -1 to +1. A positive correlation coefficient indicates a positive relationship, while a negative correlation coefficient indicates a negative relationship. The magnitude of the correlation coefficient represents the strength of the relationship, with values closer to 1 indicating a stronger correlation. From the given options, the correlation coefficient is 0.8398, indicating a strong positive correlation between the statistics of practice games and official games. Therefore, the correct answer is A. 0.8398.

(c) The alternate hypothesis in a hypothesis test represents the claim or statement that the researcher is trying to support or prove. In this case, the null hypothesis (\(H_0\)) assumes that the correlation coefficient is equal to zero (no relationship), while the alternate hypothesis (\(H_1\)) assumes that the correlation coefficient is not equal to zero (there is a relationship). From the given options, the correct alternate hypothesis is D. \(H_1: \rho \neq 0\).

(d) The test statistic is calculated to assess the strength of evidence against the null hypothesis. In this case, the test statistic is used to determine whether the correlation coefficient is significantly different from zero. The specific test statistic used for this purpose is typically the t-statistic. From the given options, the correct test statistic is C. 4.794.

(e) The degrees of freedom represent the number of independent pieces of information available for estimating a statistic. In the context of a correlation coefficient, the degrees of freedom are calculated as the sample size minus 2. Since the sample size is 12, the correct answer is A. 11.

(f) The conclusion of a hypothesis test is based on comparing the calculated test statistic with the critical value at a chosen significance level (usually 5%). If the calculated test statistic falls within the rejection region, we reject the null hypothesis in favor of the alternate hypothesis. In this case, if there is evidence to suggest that the correlation coefficient is significantly different from zero, it means that there is a relationship between the statistics of practice games and official games. From the given options, the correct conclusion is C. not zero.

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The slope of the given line is expressed as: Slope = 2

The slope of a line is defined as a measure of its steepness. Mathematically, the line slope is calculated as "rise over run" that is (change in y divided by change in x).

The formula for slope between two coordinates is expressed as:

Slope = (y₂ - y₁)/(x₂ - x₁)

The two coordinates we will use from the graph are:

(4, 0) and (0, -8)

Thus:

Slope = (-8 - 0)/(0 - 4)

Slope = 2

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Answer:

missing side.

AB²=AC²+CB²

AC²= 5²+4²

AC = √41

The correct option is d. 0.0228.  To find the p-value, we first need to calculate the test statistic:

t = (x - μ) / (σ / sqrt(n))

Where x is the sample mean, μ is the hypothesized population mean, σ is the population standard deviation, and n is the sample size.

In this case:

x = 31.1

μ = 32

σ = 2.7

n = 36

So,

t = (31.1 - 32) / (2.7 / sqrt(36)) = -2.53

Next, we need to find the p-value associated with this test statistic using a t-distribution table or calculator with 35 degrees of freedom (df = n-1).

Using a two-tailed test and α = 0.05, the p-value is approximately 0.0228.

Therefore, the correct option is d. 0.0228

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It is given that the number of miles driven by a truck driver falls between 375 and 725, and follows a uniform distribution. The probability that x lies between a and b is given by: Therefore, the probability that x < 560 is 64.29%.

The value of a is 375 and 725

Given: The number of miles driven by a truck driver falls between 375 and 725, and follows a uniform distribution.

It is given that the number of miles driven by a truck driver falls between 375 and 725.

Therefore, the value of a is 375.2.

The value of b is 725.

So, the probability density function of uniform distribution is as follows:

We have to find h.

For that, we can use the following formula:

Therefore, the value of h is 0.0029 (4 d.p).4.

The mean time to fix the furnace is 550

The mean of uniform distribution is given by:

Therefore, the mean time to fix the furnace is 550.5.

The standard deviation of time to fix the furnace is 6.

Probability P(x < 560) = 64.29

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Therefore, the 98% confidence interval for the population mean μ is approximately (23.96, 28.44).

To find the 98% confidence interval for the population mean μ, we can use the formula:

Confidence Interval = (sample mean) ± (critical value) * (standard deviation / √(sample size))

First, we need to determine the critical value associated with a 98% confidence level. Since the sample size is large (n = 411), we can use the Z-table to find the critical value. For a 98% confidence level, the critical value is approximately 2.326.

Plugging in the given values into the formula, we have:

Confidence Interval = 26.2 ± 2.326 * (21.1 / √(411))

Calculating the standard error (standard deviation / √(sample size)):

Standard Error = 21.1 / √(411)

≈ 1.04

Substituting the values:

Confidence Interval = 26.2 ± 2.326 * 1.04

Calculating the upper and lower bounds of the confidence interval:

Upper bound = 26.2 + (2.326 * 1.04)

≈ 28.44

Lower bound = 26.2 - (2.326 * 1.04)

≈ 23.96

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To find the Taylor series about 0 for the function f(x) = x² sin(x²), we can use the Maclaurin series expansion of sin(x): sin(x) = x - (1/3!)x³ + (1/5!)x⁵ -...

Substituting x² for x in the above series, we get: sin(x²) = x² - (1/3!)(x²)³ + (1/5!)(x²)⁵ - ... Multiplying by x², we have: x² sin(x²) = x⁴ - (1/3!)(x²)⁴ + (1/5!)(x²)⁶ - ... The first three non-zero terms in the Taylor series are: x² sin(x²) = x⁴ - (1/3!)x⁶ + (1/5!)x⁸ + ... The general term in the series can be written as: aₙ = (-1)^(n+1) * (1/(2n+1)!) * x^(2n+4) B. For the function g(x) = 2 cos(x) + x², the Taylor series about 0 is: cos(x) = 1 - (1/2!)x² + (1/4!)x⁴ - ... Multiplying by 2 and adding x², we get: 2 cos(x) + x² = 2 + (1 - (1/2!)x² + (1/4!)x⁴ - ...) + x². Simplifying, we have: 2 cos(x) + x² = 2 + x² - (1/2!)x² + (1/4!)x⁴ + ... The first three non-zero terms in the Taylor series are: 2 cos(x) + x² = 2 + x² - (1/2!)x² + (1/4!)x⁴ + ... The general term in the series can be written as: bₙ = (-1)ⁿ * (1/((2n)!)) * x^(2n). For the Taylor polynomial of degree 3 around the point x = -2 for f(x) = 4 + x, we need to find the values of f, f', f'', and f''' at x = -2. f(-2) = 4 + (-2) = 2; f'(-2) = 1; f''(-2) = 0; f'''(-2) = 0. Using these values, we can write the Taylor polynomial of degree 3 as: P₃(x) = f(-2) + f'(-2)(x + 2) + (f''(-2)/2!)(x + 2)² + (f'''(-2)/3!)(x + 2)³ = 2 + 1(x + 2) + 0(x + 2)² + 0(x + 2)³ = x + 4. For the function f(x) = 73 * cos(x), we can build a Maclaurin series for cos(x) and evaluate the 9th derivative of f at x = 0. The Maclaurin series for cos(x) is: cos(x) = 1 - (1/2!)x² + (1/4!)x⁴ - (1/6!)x⁶ + ... The 9th derivative of cos(x) is: cos⁽⁹⁾(x) = (1/7!)(7!) = 1.

Since f(x) = 73 * cos(x), the 9th derivative of f at x = 0 is also 1. Therefore, f⁽⁹⁾(0) = 1.

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Without the accompanying table or data, it is not possible to determine the relationship between the sweetness index and the amount of water-soluble pectin in orange juice or perform a simple linear regression to predict sweetness from pectin.

To determine whether there is a relationship between the sweetness index and the amount of water-soluble pectin in orange juice, we can use simple linear regression.

This analysis helps us understand the strength and direction of the relationship, as well as the ability to predict sweetness based on the amount of pectin.

The data collected on these two variables are shown in the accompanying table, which is not provided in the question.

Please provide the table or the relevant data so that we can proceed with the analysis and regression.

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There is a relationship between the sweetness index and the amount of water-soluble pectin in orange juice. The sweetness index of the orange juice can be predicted from the amount of pectin in parts per million.

The regression line can be computed by the method of least squares. The regression equation is `y = mx + c`,

where `y` is the dependent variable and `x` is the independent variable.The value of `m`, the slope of the regression line, is given by `m = Σ[(xi - x)(yi - y)]/Σ(xi - x)^2`.

Putting the values in the formula, we have `m = [(50-92.5)(55-85.5) + (60-92.5)(60-85.5) + ... + (160-92.5)(135-85.5)]/[(50-92.5)^2 + (60-92.5)^2 + ... + (160-92.5)^2]`.

On simplification, we have `m = 0.9124`.Therefore, the equation of the regression line is `y = 0.9124x + c`.The value of `c`, the intercept of the regression line, is given by `c = y - mx`.

Putting the values in the formula, we have `c = 85.5 - (0.9124)(92.5)`.

On simplification, we have `c = -4.33`.

Therefore, the equation of the regression line is `y = 0.9124x - 4.33`.

To predict the sweetness of the orange juice from the amount of pectin, substitute the values of `x` in the regression equation.

For example, if the amount of pectin is `100 parts per million`, then the predicted value of the sweetness index is `y = 0.9124(100) - 4.33 = 87.17`.

Therefore, the predicted value of the sweetness index is `87.17`.

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1. The probability of getting exactly two Heads when tossing three coins is not 1/2.

2. The probability that both X and Y are selected is 8/35.

3. The distribution is most likely negatively skewed based on the given mean, median, and mode.

The probability of getting exactly two Heads when tossing three coins is not 1/2. When we examine all the possible outcomes of tossing three coins (HHH, HHT, HTH, THH, TTH, THT, HTT, and TTT), we find that only three of them have exactly two Heads. Therefore, the probability of getting exactly two Heads is 3/8, not 1/2.

The probability that both X and Y are selected for the job can be calculated by multiplying their individual probabilities of selection. X has a probability of 2/5 of being selected, and Y has a probability of 4/7. Multiplying these probabilities gives us 8/35, which represents the probability that both X and Y are selected.

The given information about the mean, median, and mode suggests that the distribution is most likely negatively skewed. The fact that the mean is higher than the median indicates the presence of some higher values in the dataset, which pulls the average up.

Additionally, the mode being lower than both the mean and median suggests a clustering of values towards the lower end. These characteristics align with a negatively skewed distribution, where the tail extends towards the left side.

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Answer: Please see attached image.

Step-by-step explanation:

Resizing an item uses a transformation called dilation. Dilation is used to enlarge or shrink figures.

A scale factor is the quantity or conversion factor that is used to increase or decrease a figure's size without altering its shape.

For example in 2., we used a scale factor of 1/2 to create a smaller triangle. You simply multiply the scale factor towards the original coordinates (x, y) of the triangle created in 1.

For the problem given, we have to find the sum of each series with the indicated accuracy. The following are the formulas to be used to calculate the sum of series

To get an estimate of the sum of the series, we need to add some of the terms in the series. We can do this as follows:1 - 1/2

0.5 is an overestimate of the sum of the first two terms.1 - 1/2 + 1/3

0.833 is an overestimate of the sum of the first three terms.1 - 1/2 + 1/3 - 1/4

0.583 is an overestimate of the sum of the first four terms.1 - 1/2 + 1/3 - 1/4 + 1/5 ≈ 0.783 is an overestimate of the sum of the first five terms

.We can see that this series is converging, so we can expect the error to be less than 1/6, which is less than 0.05. Thus, the value of the given series with an error less than 0.05 is as follows:

Σ(-1^n-1 * 1/n) = 1 - 1/2 + 1/3 - 1/4 + .......We need to find the value of (-1 * (1/2)n), where n is from 0 to 3. So, we have to plug the given values in the above series to get the sum as follows

;Σ(-1 * (1/2)^n)

= -1/1 + 1/2 - 1/4 + 1/8

We can see that this series is converging, so we can expect the error to be less than 1/16, which is less than 0.005. Thus, the value of the given series with an error less than 0.005 is as follows:

Σ(-1 * (1/2)^n) = -1/1 + 1/2 - 1/4 + 1/8

The answer to the given problem is (-1n-1 * 1/n) = 0.694.

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In this problem, we are given the initial concentration of insecticide in a pond and the rates at which polluted water enters and exits the pond. We need to determine the amount of insecticide in the pond over time, write a differential equation to model this situation, analyze the long-term behavior of the insecticide concentration, and find the time it takes for the concentration to reach a critical threshold.

(a) Let's denote the amount of insecticide in the pond at time t as y(t). The rate of change of y(t) is influenced by two factors: the inflow of polluted water and the outflow of water from the pond. The differential equation that governs this situation is dy/dt = (10 g/m³ - y(t)/2100 m³) * 27 m³/day. The initial condition is y(0) = 3.5 g/m³.

(b) To analyze the long-term behavior, we need to find the equilibrium point of the differential equation. As time goes to infinity, the concentration will approach the concentration of the inflowing water, which is 10 g/m³. Therefore, the concentration of insecticide in the pond will eventually stabilize at 10 g/m³.

(c) To find the time it takes for the insecticide concentration to reach 8 g/m³, we solve the differential equation with the initial condition and track the time until y(t) reaches 8 g/m³. The exact solution will depend on the specific form of the differential equation.

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a. The expected age of a person chosen at random from the population is 47.0000 years.

b. The probability that a randomly chosen person will be older than 59 is 0.3043.

c. The 25th percentile for ages is 34.7500 years.

d. No, the Central Limit Theorem cannot be used to find the probability that the average of a sample of 12 people is less than 50, as the sample size is less than 30 and the population distribution is not normal.

e. The probability that the mean age of a random sample of 36 people is less than 50 is 0.0000.

f. The cutoff for the top 25% of mean ages, when taking repeated samples of 36 people, is 55.1667 years.

The expected age can be calculated by taking the average of the minimum and maximum ages in the uniform distribution. In this case, (20 + 74) / 2 = 47.0000 years.

To find the probability that a person will be older than 59, we need to calculate the proportion of the population that falls in the range of 59 to 74 years. Since the distribution is uniform, this is equal to (74 - 59) / (74 - 20) = 0.3043.

The 25th percentile represents the value below which 25% of the data falls. In this case, we can find the age that corresponds to the cumulative probability of 0.25 in the uniform distribution. This can be calculated as (0.25 * (74 - 20)) + 20 = 34.7500 years.

The Central Limit Theorem states that for large sample sizes (typically n ≥ 30) and regardless of the shape of the population distribution, the sampling distribution of the sample mean tends to follow a normal distribution. However, in this case, the sample size is less than 30, so the Central Limit Theorem cannot be used.

Similar to part d, the probability that the mean age of a sample of 36 people is less than 50 cannot be calculated using the Central Limit Theorem since the sample size is less than 30. Therefore, the probability is 0.0000.

To find the cutoff for the top 25% of mean ages when taking repeated samples of 36 people, we need to determine the value below which 75% of the sample means fall. This can be found by calculating the mean age corresponding to the cumulative probability of 0.75 in the original uniform distribution. Using a statistical calculator, we can find that this value is 55.1667 years.

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