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

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step-by-step explanation:

answer: solve 5\cdot 2^x=2405⋅2  

x

=2405, dot, 2, start superscript, x, end superscript, equals, 240.

solution

to solve for xxx, we must first isolate the exponential part. to do this, divide both sides by 555 as shown below. we do not multiply the 555 and the 222 as this goes against the order of operations!

\begin{aligned} 5\cdot 2^x& =240 2^x& =48 \\ \end{aligned}  

5⋅2  

x

2  

x

​    

=240

=48

​

step-by-step explanation: