Question

Weights and heights of turkeys tend to be correlated. For a population of turkeys at a farm, this correlation is found to be 0.64. The average weight is 17 pounds, SD is 5 pounds. The average height is 28 inches and the SD is 8 inches. Weight and height both roughly follow the normal curve. For each part below, answer the question or if not possible, indicate why not. A turkey at the farm which weighs more than 90% of all the turkeys is predicted to be taller than % of them. The average height for turkeys at the 90th percentile for weight is Of the turkeys at the 90th percentile for weight, roughly what percent would you estimate to be taller than 28 inches?

Answer 1

a turkey at the farm which weighs more than 90% of all the turkeys is predicted to be taller than 79.37 % of them.

The average height for turkeys at the 90th percentile for weight is 34.554

Of the turkeys at the 90th percentile for weight, roughly the percentage that would be taller than 28 inches 79.37%

Explanation:

Given that:

For a population of turkeys at a farm, the correlation found between the weights and heights of turkeys is r = 0.64

the average weight in pounds
`\overline x` = 17

the standard deviation of the weight in pounds
`S_x` = 5

the average height in inches
`\overline y` = 28

the standard deviation of the height in inches
`S_y` = 8

Also, given that the weight and height both roughly follow the normal curve

For this study , the slope of the regression line can be expressed as :

`\beta_1 = r * ( (S_y)/(S_x))`

`\beta_1 = 0.64 * ( (8)/(5))`

`\beta_1 = 0.64 * 1.6`

`\beta_1 = 1.024`

To the intercept of the regression line, we have the following equation

`\beta_o = \overline y - \beta_1 \overline x`

replacing the values:

`\beta_o = 28 -(1.024)(17)`

`\beta_o = 28 -17.408`

`\beta_o = 10.592`

However, the regression line needed for this study can be computed as:

`\hat Y = \beta_o + \beta_1 X`

`\hat Y = 10.592 + 1.024 X`

Recall that;

both the weight and height roughly follow the normal curve

As such, the weight related to 90th percentile can be determined as shown below.

Using the Excel Function at 90th percentile, which can be computed as:

(=Normsinv (0.90) ; we have the desired value of 1.28

∴

`(X - \overline x)/(s_x ) = 1.28`

`(X - 17)/(5) = 1.28`

`X - 17 = 6.4`

X = 6.4 + 17

X = 23.4

The predicted height
`\hat Y = 10.592 + 1.024 X`

where; X = 23.4

`\hat Y = 10.592 + 1.024 (23.4)`

`\hat Y = 10.592 + 23.9616`

`\hat Y = 34.5536`

Now; the probability of predicted height less than 34.5536 can be computed as:

`P(Y < 34.5536) = P( (Y - \overline y )/(S_y) < (34.5536-28)/(8))`

`P(Y < 34.5536) = P(Z< (6.5536)/(8))`

`P(Y < 34.5536) = P(Z< 0.8192)`

From the Z tables;

P(Y < 34.5536) =0.7937

Hence, a turkey at the farm which weighs more than 90% of all the turkeys is predicted to be taller than 79.37 % of them.

The average height for turkeys at the 90th percentile for weight is :

`\hat Y = 10.592 + 1.024 X`

where; X = 23.4

`\hat Y = 10.592 + 1.024 (23.4)`

`\hat Y = 10.592 + 23.962`

`\mathbf{\hat Y = 34.554}`

Of the turkeys at the 90th percentile for weight, roughly what percent would you estimate to be taller than 28 inches?

i.e

P(Y >28) = 1 - P (Y< 28)

`P(Y >28) = 1 - P( Z < (28 - 34.554)/(8))`

`P(Y >28) = 1 - P( Z < (-6.554)/(8))`

`P(Y >28) = 1 - P( Z < -0.8193)`

From the Z tables,

`P(Y >28) = 1 - 0.2063`

`\mathbf{P(Y >28) = 0.7937}`

= 79.37%