The lengths of pregnancies in a small rural village are normally distributed with a mean of 267 days and a standard deviation of 15 days. A distribution of values is normal with a mean of 267 and a standard deviation of 15. What percentage of pregnancies last beyond 246 days? P(X > 246 days) =

91.92% of pregnancies last beyond 246 days

Step-by-step explanation:

When the distribution is normal, we use 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 question, we have that:

What percentage of pregnancies last beyond 246 days?

We have to find 1 subtracted by the pvalue of Z when X = 246. So

has a pvalue of 0.0808

1 - 0.0808 = 0.9192

91.92% of pregnancies last beyond 246 days

the answer is d

step-by-step explanation:

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step-by-step explanation:

sorry no step by step