What is 1/5 X 4 ? please show simplest form
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Answer:
a turkey at the farm which weighs more than 90% of all the turkeys is predicted to be taller than <u>79.37 %</u> 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%
Step-by-step 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 = 17
the standard deviation of the weight in pounds = 5
the average height in inches = 28
the standard deviation of the height in inches = 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 :
To the intercept of the regression line, we have the following equation
replacing the values:
However, the regression line needed for this study can be computed as:
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 = 6.4 + 17
X = 23.4
The predicted height
where; X = 23.4
Now; the probability of predicted height less than 34.5536 can be computed as:
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 <u>79.37 %</u> of them.
The average height for turkeys at the 90th percentile for weight is :
where; X = 23.4
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)
From the Z tables,
= 79.37%