The appropriate feature engineering step for binning would be:
a. When we want to create defined groups from a continuous feature.
Binning is a useful technique in feature engineering when we want to convert a continuous feature into discrete or categorical groups. It involves dividing the range of values of a continuous feature into bins or intervals and assigning each value to a corresponding bin. This allows us to create defined groups or categories based on the values of the continuous feature.
Binning can be beneficial in various scenarios. For instance, it can help simplify complex data patterns, handle outliers or noise, and capture non-linear relationships between the feature and the target variable. Binning can also be used to address issues related to model complexity, data sparsity, or limited sample sizes.
By transforming a continuous feature into discrete groups, binning can enable models to capture patterns and make predictions based on the created categories. It allows for a more interpretable representation of the data and can improve the performance of certain machine learning algorithms, especially those that work better with categorical or ordinal data.
In summary, binning is an appropriate feature engineering step when we want to create defined groups or categories from a continuous feature. It can help simplify complex data patterns, handle outliers, and capture non-linear relationships, ultimately enhancing the modeling and prediction capabilities of machine learning algorithms.
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3.Water flows in a rectangular open channel with a width of 4 m. The depth of the flow is 2 m with a discharge of 20 m^3/sec. Determine the change in depth if the channel width is increased by 1 m? (negletting losses) and find the value of Fr from both conditions and identify the type of flow.
From the question above, :Width of the channel = 4 m
Depth of the flow = 2 m
Discharge = 20 m³/sec
Change in channel width = 1 mAs the discharge is constant, we have the equation of continuity as;
Q = A₁V₁ = A₂V₂
Where,Q = Discharge in m³/s
V₁ = Velocity of fluid in the original channel
A₁ = Area of the flow in the original channel
V₂ = Velocity of fluid in the new channel
A₂ = Area of the flow in the new channel
As the channel is rectangular in shape, we can write the equation as;
Q = W₁ * D₁ * V₁ = W₂ * D₂ * V₂
Where, D₁ = D₂ = D (Depth of flow)
W₁ = 4 m
W₂ = 4 + 1 = 5 m
V₁ = Velocity of fluid in the original channel
V₂ = Velocity of fluid in the new channel∴
4 * 2 * V₁ = 5 * D₂ * V₂∴ 8V₁ = 5D₂V₂∴ V₂ = 8/5 * V₁
The velocity of fluid in the new channel = 1.6V₁
Type of flow in the original channel can be determined using the Froude number as;
Fr = V₁/ √gD
Where g is the acceleration due to gravity
Fr₁ = V₁/ √gD = V₁/ √(9.81 * 2) = V₁/ 4.429
Fr₂ = V₂/ √ gD = (1.6V₁)/ √(9.81 * 2) = 1.6V₁/ 4.429
Fr₁ = V₁/ √gD = V₁/ 4.429
Fr₂ = 1.6V₁/ 4.429
Fr₁ < 1 → Subcritical flow
Fr₂ < 1 → Subcritical flow
As the value of Fr is less than 1, the type of flow is Subcritical flow
.Change in depth of the flow;
D₂ - D₁ = (W₁/W₂ - 1) * D₁D₂ - 2 = (4/5 - 1) * 2
D₂ - 2 = -0.4∴ D₂ = 1.6 m
Change in depth of the flow = D₂ - D₁ = 1.6 - 2 = -0.4 mA
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Engineers want to design seats in commercial aircraft so that they are wide enough to fit 99% of all adult would be much too expensive.) Assume adults have hip widths that are normally distributed with a mean width for adults that separates the smallest 99% from the largest 1%. What is the maximum hip width that is required to satisfy the requirement of fitting 99% of adults? in. (Round to one decimal place as needed.) Wenough to fit 99% of all adults. (Accommodating 100% of adults would require very whde seats that e normally distributed with a mean of 14.8 in. and a standard deviation of 1.1 in. Find p99. That is, find the hip of fitting 99% of adults? wdth for adults that separates the smallest 99% from the largest 1% What is the maximum hip width that is required to satisfy the requirement of fitting 99% of adults? in. (Round to one decimal place as needed)
The maximum hip width required to satisfy the requirement of fitting 99% of adults is approximately 17.0 inches.
To determine the maximum hip width needed to accommodate 99% of adults, we can use the concept of the standard normal distribution. The mean hip width for adults is given as 14.8 inches, and the standard deviation is 1.1 inches. Since we are interested in the largest 1% of hip widths, we need to find the z-score that corresponds to the cutoff point.
Using a standard normal distribution table or a calculator, we can find the z-score that represents the area of 0.99 to the left. This z-score is approximately 2.33. Now, we can use the formula z = (x - μ) / σ, where z is the z-score, x is the observed value, μ is the mean, and σ is the standard deviation.
Rearranging the formula to solve for x, we have x = z * σ + μ. Plugging in the values, x = 2.33 * 1.1 + 14.8, we get x ≈ 17.0 inches. Therefore, the maximum hip width required to accommodate 99% of adults is approximately 17.0 inches.
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