When describing categorical data, you can use: counts and proportions measures of center, spread, and shape All of these statements are correct. box plot None of these statements are correct.

Answer:

f(x) = 10 + 2.50x

Step-by-step explanation:

Let the total hours played be 'x' hours.

To find the charge for playing 'x' hours, multiply x by 2.50.

Charge for an hour = $2.50

Charge for 'x' hours = 2.50*x

                                  = $ 2.50x

To find the total cost, add one time charge with the cost charged for playing ''x'' hours.

Total cost = one-time charge + charge for 'x' hours

           f(x) = 10 + 2.50x    

x = 5 hours,

f(5) = 10 + 2.50 * 5

     = 10 + 12.50

     = $ 22.50

They have to pay $ 22.50 for playing 5 hours.

The 90% confidence interval for (p1 - p2) is (0.062, 0.178).Using a standard normal distribution table, the z-score corresponding to a 90% confidence level is approximately 1.645.

To construct a 90% confidence interval for (p1 - p2), we can use the formula:

(p1 - p2) ± zsqrt((p1(1-p1)/n1) + (p2*(1-p2)/n2))

where p1 and p2 are the sample proportions, n1 and n2 are the sample sizes, and z is the z-score corresponding to the desired confidence level.

a. n1 = 400, p^1 = 0.67, n2 = 400, p^2 = 0.55

The point estimate for (p1 - p2) is p^1 - p^2 = 0.67 - 0.55 = 0.12.

Using a standard normal distribution table, the z-score corresponding to a 90% confidence level is approximately 1.645.

Plugging in the values given, we get:

(0.67 - 0.55) ± 1.645sqrt((0.67(1-0.67)/400) + (0.55*(1-0.55)/400))

= (0.12) ± 0.058

Therefore, the 90% confidence interval for (p1 - p2) is (0.062, 0.178).

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The integral that needs to be evaluated is given by:∫36f(x)(25 – 10)30 dxTo solve the above integral, the following substitution can be made:u = x5 – 10Differentiating both sides of the above equation with respect to x gives:du/dx = 5x4Integrating both sides of the above equation with respect to x gives:dx = du/5x4

Substituting the above equation in the original integral,

we get:∫36f(x)(25 – 10)30 dx= ∫36f(x) (25 – u)30 (dx/5x4)Integrating both sides of the above equation with respect to u, we get:(1/5) ∫36f(x) (25 – u)30

Substituting the value of u back in the above equation,

we get:(1/5) ∫36f(x) (25 – (x5 – 10))30 dx(1/5) ∫36f(x) (35 – x5)30 dx

This can now be integrated using the power rule of integration to give:(1/5) * (36/6) * (35x – (1/6)x6) + C, where C is the constant of integration

Therefore, the final answer is: (6/5) * (35x – (1/6)x6) + C, where C is the constant of integration.

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In this scenario, we are dealing with a hypergeometric probability distribution with n=3, R=4, and N=9. We are tasked with calculating several probabilities and the mean and standard deviation of the distribution.

a) To calculate P(x=0), we use the formula P(x=k) = (C(R,k) * C(N-R,n-k)) / C(N,n), where C(n,r) represents the combination of choosing r items from a set of n. In this case, we have n=3, R=4, and N=9. Therefore, P(x=0) = (C(4,0) * C(9-4,3-0)) / C(9,3).

b) To calculate P(x>1), we need to calculate the probabilities for x=2 and x=3 and add them together. P(x>1) = P(x=2) + P(x=3).

c) To calculate P(x<3), we need to calculate the probabilities for x=0, x=1, and x=2 and add them together. P(x<3) = P(x=0) + P(x=1) + P(x=2).

d) The mean of the hypergeometric distribution is given by μ = n * (R/N). The standard deviation is given by σ = sqrt(n * (R/N) * (1 - R/N) * ((N-n)/(N-1))).

By substituting the given values into the formulas, we can calculate the required probabilities, mean, and standard deviation.

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The maximum likelihood estimate for μ, the mean height of the population, can be found by taking the average of the observed heights in the random sample. Given the heights: 160, 164, 150, 160, 155, 165, 170, 162, 174, and 150, we calculate their mean as follows:

(160 + 164 + 150 + 160 + 155 + 165 + 170 + 162 + 174 + 150) / 10 = 1600 / 10 = 160

Therefore, the maximum likelihood estimate for μ is 160 cm.

To find the value of fx(160), we consider the random variable X, which follows a normal distribution with mean μ and standard deviation 1. In this case, we have μ = 160, so X ~ Normal(160, 1).

The value of fx(160) represents the probability density function (PDF) of X at the value 160. Since X is normally distributed, we can use the formula for the PDF of a normal distribution to calculate this probability. However, we need the value of the standard deviation to compute the exact probability density. The question provides the standard deviation for the population heights (40), but not for the random variable X. Without the standard deviation of X, we cannot calculate the value of fx(160).

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The formula for the confidence interval of a population mean µ when the population standard deviation σ is known, is given by the following formula.

The error of estimation is the radius of the confidence interval. That is half the width of the confidence interval.Error of Estimation = z (α/2) * σ / √n Solving for the sample size, Substitute the given values,

α = 1 - 0.9 = 0.1 (since the level of confidence).

Substituting these values in the equation above,n = [1.645 * 9500 / 3000]²n ≈ 25.77, which rounds up to

n = 26.

Therefore, a sample size of at least 26 is required for the bound on the error of estimation of the 90% confidence interval to be 3000.

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(a) The 95% confidence interval for the population mean is (4.422, 6.578).

(b) My favorite homework problem this term was the one where we had to calculate the confidence interval for the population mean. I liked this problem because it was a good application of the central limit theorem, and it also required us to use our knowledge of standard deviation.

(c) I am happy with my performance in this class. I have learned a lot about statistics, and I feel confident that I can use this knowledge in my future career. I would like to thank you for your teaching, and I hope to take another statistics class with you in the future.

confidence interval = (sample mean - margin of error, sample mean + margin of error)

Plugging in the values from the problem, we get the following confidence interval:

confidence interval = (5.5 - 1.96 * 3.3 / sqrt(36), 5.5 + 1.96 * 3.3 / ✓(36))

confidence interval = (4.422, 6.578)

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The minimum recommended temperature for the cold temperature run (\#7) is 15 degrees below room temperature.

The cold temperature run, marked as \#7, requires a specific temperature range to ensure optimal performance and safety. The minimum recommended temperature for this run is 15 degrees below room temperature.

Running at temperatures below room temperature allows for a more controlled environment that mimics colder conditions, which can be beneficial for various purposes such as testing equipment, evaluating performance, or assessing durability in colder climates. It helps identify potential issues or limitations that may arise in colder environments.

Setting the minimum recommended temperature at 15 degrees below room temperature provides a sufficient cold environment without excessively low temperatures that could pose risks or potential damage.

This temperature range strikes a balance between achieving the desired cold conditions for testing purposes while ensuring the safety and integrity of the equipment or system being evaluated.

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The null hypothesis is given as follows:

[tex]H_0: \mu = 40[/tex]

The alternative hypothesis is given as follows:

[tex]H_1: \mu < 40[/tex]

The test statistic is given as follows:

t = -2.63.

The p-value is given as follows:

0.0045.

The conclusion is given as follows:

As the p-value of the test is less than the significance level of 0.05, there is enough evidence to believe that the new policies were effective in reducing wait time, which have practical significance.

The equation for the test statistic is given as follows:

[tex]t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}[/tex]

In which:

The parameters for this problem are given as follows:

[tex]\overline{x} = 39.3, \mu = 40, s = 4.2, n = 250[/tex]

Hence the test statistic is given as follows:

[tex]t = \frac{39.3 - 40}{\frac{4.2}{\sqrt{250}}}[/tex]

t = -2.63.

Using a t-distribution calculator, for a left-tailed test, as we are testing if the mean is less than a value, with t = -2.63 and 250 - 1 = 249 df, the p-value is given as follows:

0.0045.

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The estimated demand equation for the quantity demanded of a sofa based on the CEO's data is Qi = 897.8 - 0.068Pi.

In this equation, Qi represents the quantity demanded of the sofa, and Pi represents the price of the sofa. The estimated equation suggests that as the price of the sofa (Pi) increases, the quantity demanded (Qi) decreases. The intercept term, 897.8, indicates the estimated quantity demanded when the price is zero. The slope term, -0.068, represents the rate of change in quantity demanded for each unit increase in price.

The reported R-squared value of 0.75 indicates that approximately 75% of the variation in quantity demanded can be explained by the linear relationship with price. The remaining 25% is attributed to other factors not accounted for in the estimated equation. Overall, the estimated equation provides a quantitative relationship between the price and quantity demanded of the sofa based on the CEO's data.

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The expected value of a discrete probability distribution with n = 8 equally equally likely outcomes is 4.

The expected value of a discrete probability distribution is the weighted average of the possible values, where the weights are the probabilities of the values occurring. In this case, there are 8 equally likely outcomes, so each outcome has a probability of 1/8. The possible values are 1, 2, 3, 4, 5, 6, 7, and 8. The expected value is therefore:

Expected value = (1/8) * 1 + (1/8) * 2 + ... + (1/8) * 8 = 4

The expected value is a measure of the central tendency of a probability distribution. It is the value that we would expect to occur if we repeated the experiment many times. In this case, we would expect to get a value of 4 on average if we rolled a die many times.

The expected value is also known as the mean. The mean is calculated by adding up all of the possible values and dividing by the number of possible values. In this case, the mean is also 4.

The expected value is a useful measure of central tendency because it is not affected by extreme values. For example, if we rolled a die and got a 1 on every roll, the mean would still be 4. This is because the expected value is a weighted average, and the probability of getting a 1 is very low, so it does not have a large impact on the average.

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a) The mean satisfaction rating for Restaurant A is 3, and for Restaurant B is 16., b) The standard deviation for Restaurant A is approximately 0.632, and for Restaurant B is approximately 7.783.

a. To calculate the mean satisfaction rating at each location, we need to find the average of the ratings for each restaurant.

For Restaurant A:

Mean = (1 + 2 + 3 + 4 + 5) / 5 = 3

For Restaurant B:

Mean = (6 + 10 + 7 + 39 + 18) / 5 = 16

The mean satisfaction rating for Restaurant A is 3, and for Restaurant B is 16.

b. To calculate the standard deviation of each distribution, we need to find the measure of how spread out the ratings are from the mean.

For Restaurant A:

Standard Deviation = [tex]\sqrt{[((1 - 3)^2 + (2 - 3)^2 + (3 - 3)^2 + (4 - 3)^2 + (5 - 3)^2) / 5]}[/tex]

= [tex]\sqrt{[2.0 / 5]}[/tex]

≈ [tex]\sqrt{0.4}[/tex]

≈ 0.632

For Restaurant B:

Standard Deviation = [tex]\sqrt{[((6 - 16)^2 + (10 - 16)^2 + (7 - 16)^2 + (39 - 16)^2 + (18 - 16)^2) / 5]}[/tex]

= [tex]\sqrt{[302.8 / 5]}[/tex]

≈ [tex]\sqrt{60.56}[/tex]

≈ 7.783

The standard deviation for Restaurant A is approximately 0.632, and for Restaurant B is approximately 7.783.

c. From these results, we can draw the following conclusions:

The mean customer satisfaction rating for Restaurant A is 3, while for Restaurant B it is 16. Therefore, Restaurant B has a higher average customer satisfaction rating compared to Restaurant A.

The standard deviation for Restaurant A is approximately 0.632, indicating that the ratings are relatively consistent and close to the mean. In contrast, the standard deviation for Restaurant B is approximately 7.783, indicating that the ratings are more spread out and less consistent compared to the mean.

Based on these conclusions, we can infer that Restaurant B generally has higher customer satisfaction ratings compared to Restaurant A, and the ratings for Restaurant A are more consistent compared to Restaurant B.

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    The correct option among the following given options that describes the central limit theorem is option D, i.e., If the distribution of a random variable is non-normal, the sampling distribution of the sample mean will be approximately normal for samples n ≥ 30.

     What is the Central Limit Theorem? The Central limit theorem states that the sum or the average of a large number of independent and identically distributed (i.i.d.) random variables approaches a normal distribution, even if the original random variable itself is not normally distributed. The central limit theorem is important because it allows statisticians to make conclusions about the population by examining a sample of data. The central limit theorem is a statistical theory that forms the basis for much of modern statistical inference. According to the central limit theorem, the sampling distribution of the sample mean will be approximately normal if the distribution of a random variable is non-normal, for samples n≥30.

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The sample standard deviation was calculated as approximately 0.727 using the range and critical value from the confidence interval.

The sample standard deviation can be calculated using the formula:

s is equal to Range by 2 times Critical Value

In this case, the range of the confidence interval is given by the difference between the upper and lower bounds: \(57.5 - 54.5 = 3\). The critical value is a measure of how many standard deviations away from the mean the interval extends, and for a 95% confidence interval with a sample size of 25, the critical value is approximately 2.060.

Substituting these values into the formula, we get:

s is equal to 3 by 2 times 2.060 is is approximately 0.727.

Therefore, the sample standard deviation is approximately 0.727.

The sample standard deviation is a measure of the dispersion or spread of data within a sample. It is calculated as the square root of the sample variance. The sample variance, in turn, is calculated by taking the average of the squared differences between each data point and the sample mean.

To calculate the sample standard deviation from a confidence interval, we need to consider the range of the interval and the critical value. The range is the difference between the upper and lower bounds of the interval. The critical value represents the number of standard deviations away from the mean the interval extends, and it depends on the desired level of confidence and the sample size.

In this case, we were given the range of the confidence interval and the fact that it corresponds to a 95% confidence level. By referring to a table or using statistical software, we can determine the critical value associated with a 95% confidence level for a sample size of 25, which is approximately 2.060.

Using the formula for sample standard deviation and substituting the range and critical value, we calculated the sample standard deviation to be approximately 0.727.

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The most appropriate chart to show the change in average global temperature over the course of the 20th century in a presentation on climate change would be a time-series chart.

A time-series chart is designed to display data points over a continuous time interval. It is commonly used to visualize trends and patterns in data that are recorded over time. In the case of tracking average global temperature, a time-series chart would be ideal as it allows for the representation of temperature data points at different time intervals throughout the 20th century.

By using a time-series chart, each data point representing the average global temperature for a specific year can be plotted along the time axis. This would provide a clear visual representation of the temperature variations over time and enable viewers to observe any upward or downward trends in temperature.

On the other hand, a bar chart is more suitable for comparing categorical data or discrete variables, a scatter plot is ideal for visualizing the relationship between two continuous variables, and a frequency polygon is typically used to display the distribution of a single variable. Therefore, for showing the change in average global temperature over time, a time-series chart is the most appropriate choice.

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The expectation for the given probability distribution is 8.9.

To calculate the expectation or mean of a probability distribution, we multiply each value of the random variable by its corresponding probability and then sum up the products. In this case, the random variable x takes the values 2, 6, 10, 14, and 18 with probabilities 0.5, 0.3, 0.1, 0.06, and 0.04 respectively.

The calculation is as follows:

E(x) = (2 * 0.5) + (6 * 0.3) + (10 * 0.1) + (14 * 0.06) + (18 * 0.04)

    = 1 + 1.8 + 1 + 0.84 + 0.72

    = 5.36

Therefore, the expectation or mean of the probability distribution is 8.9.

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In the given scenario, Machine 1 has produced 712 satisfactory and 178 unsatisfactory pistons.

Machine 2 has produced 486 satisfactory and 414 unsatisfactory pistons.

[tex]Therefore, the total number of pistons produced by Machine 1 and Machine 2 are 712+178=890 and 486+414=900, respectively.[/tex]

[tex]The probability of choosing a satisfactory piston from Machine 1 is 712/890.[/tex]

The probability of choosing an unsatisfactory piston from Machine 2 is 414/900.

The probability of choosing a satisfactory piston from Machine 1 and an unsatisfactory piston from Machine 2 is the product of the above two probabilities[tex]:712/890 × 414/900 = 0.329[/tex]

Therefore, the probability that the piston chosen from Machine 1 is satisfactory and the piston chosen from Machine 2 is unsatisfactory is 0.329 or approximately 0.33.

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1. Probability of winning the jackpot = (11,238,513 x 26) / (C(69,5) x 26) ≈ 1 in 292,201,338

2. Playing this lottery is not a good investment if your goal is to make money.

The probability of winning the jackpot from one ticket purchase is calculated by multiplying the probability of choosing 5 correct numbers out of 69 and the probability of choosing the correct powerball number out of 26. So, we have:

Probability of winning the jackpot = (Number of ways to choose 5 correct numbers out of 69) x (Number of ways to choose 1 correct powerball number out of 26) / (Total number of possible combinations)

The number of ways to choose 5 correct numbers out of 69 is given by the combination formula C(69,5), which equals 11,238,513. The number of ways to choose 1 correct powerball number out of 26 is simply 26. The total number of possible combinations is given by the product of the total number of ways to choose 5 numbers out of 69 and the total number of ways to choose 1 powerball number out of 26, which is equal to C(69,5) x 26.

Therefore, the probability of winning the jackpot from one ticket purchase is:

Probability of winning the jackpot = (11,238,513 x 26) / (C(69,5) x 26) ≈ 1 in 292,201,338

To find the expected value of your winnings from playing this lottery, we need to multiply the probability of winning by the amount you would win and subtract the cost of purchasing a ticket.

So, the expected value of your winnings is:

Expected value = (Probability of winning) x (Amount you would win) - (Cost of ticket)

Expected value = (1/292,201,338) x ($40,000,000) - ($2) ≈ -$1.37

This means that on average, for every $2 spent on a ticket, you can expect to lose about $1.37. Therefore, playing this lottery is not a good investment if your goal is to make money.

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Answer:

[tex] \sqrt[3]{1000} = 10[/tex]

So the radius of the sphere is 10/2 = 5 inches.

The volume of this sphere is

(4/3)π(5³) = 500π/3 cubic inches.

The 99 percent confidence interval for the population mean is (7.351, 12.649) gallons.

To compute a 99 percent confidence interval for the population mean, we can use the formula:

Confidence Interval = Sample Mean ± Margin of Error

The margin of error is calculated as:

Margin of Error = Critical Value c (Population Standard Deviation / √Sample Size)

Since the sample size is large (n > 30), we can use the Z-distribution. The critical value for a 99 percent confidence level is 2.576.

Next, Margin of Error = 2.576 (5.10 / √60) ≈ 2.649

Finally, we can construct the confidence interval:

Confidence Interval = 10.00 ± 2.649

The 99 percent confidence interval for the population mean is (7.351, 12.649) gallons.

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The standard error of the mean for the sample of eight is approximately 3.16.

To calculate the standard error of the mean (SEM), we can use the formula:

SEM = standard deviation / square root of sample size

For the sample of four:

Standard deviation (s) = 5.48

Sample size (n) = 4

SEM = 5.48 / sqrt(4) = 5.48 / 2 = 2.74

Therefore, the standard error of the mean for the sample of four is 2.74.

For the sample of eight:

Standard deviation (s) = 8.93

Sample size (n) = 8

SEM = 8.93 / sqrt(8) ≈ 3.16

Therefore, the standard error of the mean for the sample of eight is approximately 3.16.

Comparing the two samples, the sample of four has a smaller standard error of the mean (2.74) compared to the sample of eight (approximately 3.16). A smaller standard error indicates a more precise estimate of the population mean. Therefore, the sample of four is a better estimate of the population mean compared to the sample of eight.

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Given that θ = tan⁻¹(-2) + πa) Find cosθ and sinθ.The given θ can be written as :θ = tan⁻¹(-2) + πθ = tan⁻¹(-2) + π(but we don't know what the value of 'a' is)In order to calculate sin θ and cos θ, we need to find the exact value of θ first.Then we have,tan θ = -2=> θ = tan⁻¹(-2) = -63.43°π radians = 180°cos (-63.43°) = 0.447 and sin (-63.43°) = -0.894Therefore,cos θ = 0.447sin θ = -0.894b) Let z=4√5cisθ. Find all cube roots of z/3-i.From the question we have, z=4√5cisθNow we need to find the cube roots of (z/3-i)Let's first find the value of z/3-i:(z/3-i) = (4√5cisθ)/(3-i)To simplify, we can multiply and divide the numerator and denominator by (3+i), which is the conjugate of (3-i).(z/3-i) = [(4√5cisθ)/(3-i)]*[(3+i)/(3+i)]= (12√5cisθ+4√5i)/10= (6√5cisθ+2√5i)/5= (2/5)(3√5cisθ+√5i)Now we have (2/5)(3√5cisθ+√5i), which is in polar form and can be written as r(cosθ + i sinθ).We know that the cube roots of any complex number in polar form r(cosθ + i sinθ) can be found by using the following formula:r1/3[cos((θ + 2πk)/3) + i sin((θ + 2πk)/3)]Where k = 0,1,2.Let's apply the above formula to find the cube roots of (z/3-i)r = 2/5 = 0.4(θ + i sin θ) = (3√5cisθ+√5i)/5So,r1/3[cos((θ + 2πk)/3) + i sin((θ + 2πk)/3)] = 0.4^(1/3){cos[((θ + 2πk)/3)] + i sin[((θ + 2πk)/3)]}k = 0=> 0.4^(1/3){cos[(θ/3)] + i sin[(θ/3)]} = 0.702 cis(20.54°)k = 1=> 0.4^(1/3){cos[(θ + 2π)/3] + i sin[(θ + 2π)/3]} = 0.702 cis(186.17°)k = 2=> 0.4^(1/3){cos[(θ + 4π)/3] + i sin[(θ + 4π)/3]} = 0.702 cis(351.79°)Therefore, the three cube roots of (z/3-i) are:0.702 cis(20.54°)0.702 cis(186.17°)0.702 cis(351.79°)Hence, the three cube roots of (z/3-i) are 0.702 cis(20.54°), 0.702 cis(186.17°), and 0.702 cis(351.79°).

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The probability that the person's score is between 42 and 48 is approximately 0.6827.

To find the probability that the person's score is between 42 and 48, we need to calculate the area under the normal distribution curve between these two values.

First, we standardize the values by subtracting the mean and dividing by the standard deviation.

For 42, the standardized score (z-score) is:

(42 - 45) / 9 = -0.3333

For 48, the standardized score (z-score) is:

(48 - 45) / 9 = 0.3333.

Next, we look up these z-scores in the standard normal distribution table or use a calculator to find the corresponding cumulative probability.

Using the standard normal distribution table, we find that the cumulative probability for a z-score of -0.3333 is approximately 0.3707, and the cumulative probability for a z-score of 0.3333 is approximately 0.6293.

Finally, we subtract the cumulative probability of the lower value from the cumulative probability of the higher value to get the probability between the two values:

P(42 ≤ x ≤ 48) = 0.6293 - 0.3707

= 0.2586

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Probabilities:

(a) P(CUD) = 0.9

(b) P(Cc n D) = 0.4

(c) P(CnD) = 0.3

In order to solve these probabilities, we need to understand the concepts of intersection and complement.

P(C) represents the probability of event C occurring, while P(D) represents the probability of event D occurring. P(CND) represents the probability of both events C and D occurring simultaneously.

(a) To find P(CUD), we need to find the probability of event C or event D occurring. This can be calculated using the formula: P(CUD) = P(C) + P(D) - P(CND). Plugging in the given values, we have P(CUD) = 0.5 + 0.6 - 0.2 = 0.9.

(b) To find P(Cc n D), we need to find the probability of the complement of event C and event D occurring simultaneously.

The complement of event C is denoted as Cc. The formula for P(Cc n D) is: P(Cc n D) = P(D) - P(CnD). Substituting the given values, we have P(Cc n D) = 0.6 - 0.2 = 0.4.

(c) Now, if events C and D are independent, then the probability of both events occurring simultaneously is equal to the product of their individual probabilities. Therefore, P(CnD) = P(C) * P(D) = 0.5 * 0.6 = 0.3.

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Hypothesis testing is an essential statistical tool that is used to make an inference about a population parameter using a sample statistic.

The following are the null and alternative hypotheses;

Null hypothesis[tex]H0: μ1 - μ2 = 0,[/tex] the average weight of one week old grey seal pups on the west side of the island is the same as the average weight of one week old grey seal pups on the east side of the island.

Alternative hypothesis[tex]H1: μ1 - μ2 ≠ 0[/tex], the average weight of one week old grey seal pups on the west side of the island is different from the average weight of one week old grey seal pups on the east side of the island.

The standard error of the test statistic (Spooled) is calculated as shown below;[tex]S_p=\sqrt{\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2}}[/tex][tex]S_p=\sqrt{\frac{(30-1)13.425^2+(24-1)13.325^2}{30+24-2}}[/tex][tex]S_p=13.375[/tex]

The test statistic is calculated as shown below;[tex]t=\frac{(\overline{x_1}-\overline{x_2})-(\mu_1-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}[/tex][tex]t=\frac{(31.697-27.250)-(0)}{13.375\sqrt{\frac{1}{30}+\frac{1}{24}}}[/tex][tex]t=1.89[/tex]The degrees of freedom for the t-distribution are [tex]f=n1+n2-2=df=30+24-2=df=52[/tex]

For a significance level α=0.1 and a two-tailed test, the critical value of t is ±1.675.

Therefore, since the calculated test statistic is not greater than the critical value of [tex]tα/2,df=52[/tex]

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The growth rate between 2015 and 2016 of this random variable was 2%.

Given that the data collection is {100, 102, 104, 106, 105, 108}.

First element 100 is the value of some random variable in year 2015.102 is the value of the random variable in year 2016.Now the growth rate between 2015 and 2016 of this random variable was (in %) .

Growth rate= (final value - initial value)/initial value×100Given initial value=100, final value=102Growth rate= (102-100)/100 × 100=2%

Therefore, the growth rate between 2015 and 2016 of this random variable was 2%.

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Summary (30 words):

a. The area under the standard normal curve between -2.13 and 0.1 is approximately 0.4452.

b. The area under the standard normal curve between -0.00999999999999979 and 2.22 is approximately 0.4854.

c. The area under the standard normal curve to the left of 1.69 is approximately 0.9554.

d. The area under the standard normal curve to the left of -1.45 is approximately 0.0735.

e. The area under the standard normal curve to the right of -1.63 is approximately 0.9484.

f. The area under the standard normal curve to the right of 0.01 is approximately 0.5040.

a. The area under the standard normal curve between -2.13 and 0.1, we need to find the corresponding cumulative probabilities for each value and subtract them. Using a standard normal distribution table or a calculator, we find that the cumulative probability for -2.13 is approximately 0.0166 and for 0.1 is approximately 0.5398. Subtracting 0.0166 from 0.5398 gives us 0.5232.

b. Similarly, to find the area under the standard normal curve between -0.00999999999999979 and 2.22, we calculate the cumulative probabilities for each value and subtract them. The cumulative probability for -0.00999999999999979 is approximately 0.4960 and for 2.22 is approximately 0.9857. Subtracting 0.4960 from 0.9857 gives us 0.4897.

c. To find the area under the standard normal curve to the left of 1.69, we look up the cumulative probability for 1.69 in the standard normal distribution table, which is approximately 0.9554.

d. To find the area under the standard normal curve to the left of -1.45, we look up the cumulative probability for -1.45, which is approximately 0.0735.

e. To find the area under the standard normal curve to the right of -1.63, we subtract the cumulative probability for -1.63 from 1. The cumulative probability for -1.63 is approximately 0.0516, so the area to the right is approximately 0.9484.

f. To find the area under the standard normal curve to the right of 0.01, we subtract the cumulative probability for 0.01 from 1. The cumulative probability for 0.01 is approximately 0.4960, so the area to the right is approximately 0.5040.

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The correct option is; A. Approximately normal because n ≤ 0.05N and np(1 - p) < 10, best describes the shape of the sampling distribution of p.What is sampling distribution?The distribution of the values of the statistic that comes from the random sampling of the population is called the sampling distribution.

For instance, the proportion of people who purchase a particular product, the mean weight of people belonging to a specific group, the difference between two population means, and so on are all statistics.

σp = √[pq/n]where:p is the population proportion (of a particular characteristic)q is 1-pn is the sample size

Therefore, here is the sampling distribution of p for the given values of n, p, and N(15000).Here,

N=15000np=600 × 0.7

= 420 (i.e., the mean of the distribution)

q=0.3

n=600

Now we can check the normality of the distribution using the following criteria:n ≤ 0.05N and np(1 - p) < 10n = 600,

N = 15000n/N

= 600/15000

= 0.04 (≤0.05)np(1 - p)

= 420(0.3)

= 126 (≤10)

Therefore, the shape of the sampling distribution of p is approximately normal because n ≤ 0.05N and np(1 - p) < 10.

The correct option is A.

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a. This is a two-tailed test.

b. P-value = 0.0436 (rounded to three decimal places).

Given:

Test statistic (z) = 1.72

Null hypothesis ([tex]H_0[/tex]): p = 0.342

Significance level (α) = 0.01

a. To identify the hypothesis test, we need to determine whether it is a two-tailed, left-tailed, or right-tailed test. Since the alternative hypothesis is stated as p ≠ 0.342, it indicates a two-tailed test. This means we are testing for deviations in both directions.

b. To find the p-value for a two-tailed test, we need to calculate the area in both tails beyond the absolute value of the test statistic.

Using a standard normal distribution table or a statistical calculator, we can find the cumulative probability associated with a z-value of 1.72. The cumulative probability is approximately 0.9564.

The area in one tail is (1 - 0.9564) / 2 = 0.0218.

Since this is a two-tailed test, the p-value is twice the area in one tail, which gives us p-value = 2 * 0.0218 = 0.0436.

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The 99% confidence interval is (1.31, 3.19).

The biologist is required to estimate the weight of all spotted lobsters on the Treasure Coast. A random sample of 20 spotted lobsters is collected to obtain this estimation. The weights of each lobster in the sample are given below. Assume that the weights of all spotted lobsters on the Treasure Coast are normally distributed.To determine the point estimate x¯ and the sample standard deviation s, we need to calculate the mean and standard deviation of the sample. Here is how to do it:

[tex]\[\text{x¯}=\frac{1}{n}\sum_{i=1}^{n} x_i\][/tex]

Where n = sample size, x i = individual value in the sample.

Using this formula, we get:

[tex]\[\text{x¯}=\frac{1}{20}(1.62+1.74+1.74+1.83+1.94+2.12+2.15+2.21+2.27+2.35+2.49+2.51+2.52+2.65+2.65+2.89+2.93+2.94+3.31+4.24)=\frac{44.98}{20}=2.249\][/tex]

Therefore, the point estimate is 2.249.To calculate the sample standard deviation, we use this formula:

[tex]\[s=\sqrt{\frac{\sum_{i=1}^{n}(x_i-\text{x¯})^2}{n-1}}\][/tex]

Using this formula, we get:

\[s=\sqrt{\frac{(1.62-2.249)^2+(1.74-2.249)^2+(1.74-2.249)^2+(1.83-2.249)^2+(1.94-2.249)^2+(2.12-2.249)^2+(2.15-2.249)^2+(2.21-2.249)^2+(2.27-2.249)^2+(2.35-2.249)^2+(2.49-2.249)^2+(2.51-2.249)^2+(2.52-2.249)^2+(2.65-2.249)^2+(2.65-2.249)^2+(2.89-2.249)^2+(2.93-2.249)^2+(2.94-2.249)^2+(3.31-2.249)^2+(4.24-2.249)^2}{20-1}}=\sqrt{\frac{18.0941}{19}}=0.9986\]

Therefore, the sample standard deviation is 0.9986.Using a 99% confidence level, the margin of error E is given by:

[tex]\[E=z_{\alpha/2}\frac{s}{\sqrt{n}}\][/tex]

Where α is the level of significance, s is the sample standard deviation, and n is the sample size. For a 99% confidence level, α = 0.01/2 = 0.005 and zα/2 = 2.576. Substituting the values, we get:

[tex]\[E=2.576\cdot\frac{0.9986}{\sqrt{20}}=0.9423\][/tex]

Therefore, the margin of error is 0.9423.The 99% confidence interval is given by:

[tex]\[\text{x¯}-E<\mu<\text{x¯}+E\][/tex]

Where μ is the population mean and E is the margin of error. Substituting the values, we get:

[tex]\[2.249-0.9423<\mu<2.249+0.9423\]\[1.3067<\mu<3.1913\][/tex]

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The 99% confidence interval for the average weight of a spotted lobster on the Treasure Coast is (47.74, 51.06).

The given data represents the weights of 20 randomly selected spotted lobsters.

Assuming that the weights of all spotted lobsters on the Treasure Coast are normally distributed, we have to determine the point estimate and sample standard deviation of the given data.

Step 1: Calculate the sample mean:

[tex]$$\overline{x} = \frac{1}{n}\sum_{i=1}^{n} x_i$$$$\overline{x} = \frac{52+46+47+53+49+51+48+50+53+49+48+46+49+55+50+47+52+50+51+47}{20}$$$$\overline{x} = 49.4$$[/tex]

Hence, the sample mean is 49.4.

Step 2: Calculate the sample standard deviation:

[tex]$$s = \sqrt{\frac{1}{n-1}\sum_{i=1}^{n}(x_i-\overline{x})^2}$$$$s = \sqrt{\frac{1}{19}\left[(52-49.4)^2 + (46-49.4)^2 + \cdots + (47-49.4)^2\right]}$$$$s = \sqrt{\frac{1}{19}(202.44)}$$$$s = 2.92$$[/tex]

Therefore, the sample standard deviation is 2.92.

Now, we will use the sample data to determine a 99% confidence interval for the average weight of a spotted lobster on the Treasure Coast.

Step 3: Calculate the margin of error:

[tex]$$E = z_{\alpha/2}\cdot \frac{s}{\sqrt{n}}$$[/tex]

Here, n = 20 and [tex]$\alpha$[/tex] = 1 - 0.99 = 0.01.

Using the standard normal table, we find the value of z at [tex]$\alpha/2[/tex] = 0.005$ to be 2.576.

Therefore, we have

$$E = 2.576 \cdot \frac{2.92}{\sqrt{20}} = 1.657$$

The margin of error is 1.657.

Step 4: Calculate the 99% confidence interval:

[tex]$$\text{Confidence interval} = \overline{x} \pm E$$$$\text{Confidence interval} = 49.4 \pm 1.657$$$$\text{Confidence interval} = (47.74, 51.06)$$[/tex]

Therefore, the 99% confidence interval for the average weight of a spotted lobster on the Treasure Coast is (47.74, 51.06).

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