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 $0.9938$.
Further Reading
- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators
