Question

The weights of steers in a herd are distributed normally. The standard deviation is 300lbs and the mean steer weight is 900lbs. Find the probability that the weight of a randomly selected steer is less than 1140lbs. Round your answer to four decimal places.

Answer 1

0.7611 = 76.11% probability that the weight of a randomly selected steer is less than 1140lbs.

Explanation:

Problems of normally distributed samples can be 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 = 900, \sigma = 300`

Find the probability that the weight of a randomly selected steer is less than 1140lbs.

This is the pvalue of Z when X = 1140. So

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

`Z = (1140 - 900)/(300)`

`Z = 0.71`

`Z = 0.71` has a pvalue of 0.7611

0.7611 = 76.11% probability that the weight of a randomly selected steer is less than 1140lbs.