Question

The mean weight of a breed of yearling cattle is 11921192 pounds. Suppose that weights of all such animals can be described by the Normal model ​N(11921192​,7878​). ​a) How many standard deviations from the mean would a steer weighing 10001000 pounds​ be? ​b) Which would be more​ unusual, a steer weighing 10001000 ​pounds, or one weighing 12501250 ​pounds? ​

a) A steer weighing 10001000 pounds is 2.462.46 standard deviations below the mean. ​(Round to two decimal places as​ needed.) ​
b) The steer weighing 1000 pounds is more unusual.

Answer 1

a) 2.46 standard deviations below the mean

b) The steer weighing 1000 pounds is more unusual.

Explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

`\mu = 1192, \sigma = 78`

a) How many standard deviations from the mean would a steer weighing 1000 pounds​ be?

This is Z when X = 1000.

`Z = (X - \mu)/(\sigma)`

`Z = (1000 - 1192)/(78)`

`Z = -2.46`

`Z = -2.46` is 2.46 standard deviations below the mean.

​b) Which would be more​ unusual, a steer weighing 1000 ​pounds, or one weighing 1250 ​pounds?

The higher the absolute value of the z-score, the more unusual the measure is.

X = 1000 pounds

`Z = (X - \mu)/(\sigma)`

`Z = (1000 - 1192)/(78)`

`Z = -2.46`

The absolute value of -2.46 is 2.46

X = 1250 pounds

`Z = (X - \mu)/(\sigma)`

`Z = (1250 - 1192)/(78)`

`Z = 0.74`

The absolute value of 0.74 is 0.74.

So the steer weighing 1000 pounds is more unusual.