Question

The average weight of full-grown beef cows is 1,470 pounds with a standard deviation of 230 pounds. if the weights are normally distributed, what is the percentile rank of a cow that weighs 1,750 pounds? (1) 89th (2) 76th (3) 49th (4) 35th

Answer 1

Answer: The percentile is 89

Step-by-step explanation:

This question can be solved using concept for t tables

In a normal distribution the curve. $\mu= 1470 , \sigma = 230, x=1750$

The relationship between z score, mean and standard deviation is given by

$x = \mu+z\sigma$

So the z value according to this is given by the formula

$1750=1470+z(230)\\ z=\frac{280}{230} \\\\z = 1.217$

From the z table we can infer that p value for z=+1.217 is 88.82

So 1750 is 89th percentile

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