Question

In a certain breed of cattle, the length of gestation has a mean of 284 days and a standard deviation is 5.5 days. What length of gestation, rounded to the nearest whole number, represents the 18th percentile?

Answer 1

Answer:

A gestation length of 279 days represents the 18th percentile.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions 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:

$Z = \frac{X - \mu}{\sigma}$

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 a certain breed of cattle, the length of gestation has a mean of 284 days and a standard deviation is 5.5 days.

This means that $\mu = 284, \sigma = 5.5$

What length of gestation, rounded to the nearest whole number, represents the 18th percentile?

This is X when Z has a pvalue of 0.18. So X when Z = -0.915.

$Z = \frac{X - \mu}{\sigma}$

$-0.915 = \frac{X - 284}{5.5}$

$X - 284 = -0.915*5.5$

$X = 279$

A gestation length of 279 days represents the 18th percentile.