Question

The length of gestation for swine is normally distributed with mean 114 days and standard deviation 0.75 day. Find the probability that a litter will be born within one day of the mean of 114.

Answer 1

Answer:

81.64% probability that a litter will be born within one day of the mean of 114.

Step-by-step 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:

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

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 this problem, we have that:

$\mu = 114, \sigma = 0.75$

Find the probability that a litter will be born within one day of the mean of 114.

114 + 1 = 115

114 - 1 = 113

This probability is the pvalue of Z when X = 115 subtracted by the pvalue of Z when X = 113.

So

X = 115

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

$Z = \frac{115 - 114}{0.75}$

$Z = 1.33$

$Z = 1.33$ has a pvalue of 0.9082

X = 113

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

$Z = \frac{113 - 114}{0.75}$

$Z = -1.33$

$Z = -1.33$ has a pvalue of 0.0918

0.9082 - 0.0918 = 0.8164

81.64% probability that a litter will be born within one day of the mean of 114.