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.

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 and standard deviation , the zscore of a measure X is given by:

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:

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

has a pvalue of 0.9082

X = 113

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.

for the first question it would be the second answer for the second question it would be the third answer and for the third question it would be the first answer

step-by-step explanation: