Answer:
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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:
[tex]\mu = 267, \sigma = 15[/tex]
What percentage of pregnancies last beyond 246 days?
We have to find 1 subtracted by the pvalue of Z when X = 246. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{246 - 267}{15}[/tex]
[tex]Z = -1.4[/tex]
[tex]Z = -1.4[/tex] has a pvalue of 0.0808
1 - 0.0808 = 0.9192
91.92% of pregnancies last beyond 246 days
