# Solved (Free): If the mean number of cigarettes smoked by pregnant women is 16 and the standard deviation is 8

#### ByDr. Raju Chaudhari

Apr 12, 2021

If the mean number of cigarettes smoked by pregnant women is 16 and the standard deviation is 8, find the probability that in a random sample of 100 pregnant women the mean number of cigarettes smoked will be more than 18.

Also find the probability that in a random sample of 100 pregnant women the mean number of cigarettes smoked will be less than 18.

### Solution

Let $X$ denote the number of cigarettes smoked by pregnant women.

Let $E(X)=\mu$, $V(X)=\sigma^2$, where $\mu = 16$, $\sigma = 8$.

A sample of $n = 100$ pregnant women are selected at random, then using central limit theorem for large $n$, $\overline{X} \sim N(\mu, \sigma^2/n)$.

So $Z=\frac{\overline{X}-\mu}{\sigma/\sqrt{n}} \sim N(0,1)$.

The probability that probability that in a random sample of 100 pregnant women the mean number of cigarettes smoked will be more than 18 is

 \begin{aligned} P(\overline{X} > 18) & =1-P(\overline{X} < 18)\\ &= 1-P\bigg(\frac{\overline{X} -\mu}{\sigma/\sqrt{n}}< \frac{18 -16}{8/\sqrt{100}}\bigg)\\ &= 1-P(Z < 2.5)\\ &= 1-0.9938\\ &=0.0062 \end{aligned}

Thus the probability that in a random sample of 100 pregnant women the mean number of cigarettes smoked will be more than 18 is $0.0062$.

The probability that in a random sample of 100 pregnant women the mean number of cigarettes smoked will be less than 18 is
 \begin{aligned} P(\overline{X} < 18) &= P\bigg(\frac{\overline{X} -\mu}{\sigma/\sqrt{n}} < \frac{18 -16}{8/\sqrt{100}}\bigg)\\ &= P(Z < 2.5)\\ &= 0.9938 \end{aligned}

Thus the probability that in a random sample of 100 pregnant women the mean number of cigarettes smoked will be less than 18 is $0.9938$.