It is known that the income of guests at an all-inclusive resort on the north coast is normally distributed with a standard deviation of $\$8000$.
Suppose a random sample of $50$ guests is taken:
a) What is the probability that the sample standard deviation of incomes is at least $\$5000\ ?$.
b) What is the probability that the standard deviation of the incomes for this sample is no more than $\$1000\ ?$.
With the questions above, I have tried to find the probability but I’ve never solved one of this without being provided with the mean, this is what I’m used to:
- First finding the z-score, $\left(5000 - \mbox{mean}\right)/\mbox{standard deviation}$.
- Then look for the value in the normal distribution table.
If $\ V\ $ is the sample variance $$ V=\frac{1}{49}\sum_{i=1}^{50}\left(X_i-\frac{1}{50}\sum_{j=1}^{50}X_j\right)^2 $$ —that is the square of the sample standard deviation $\ S\ $—then the random variable $$ \frac{49V}{8000^2}=\frac{49S^2}{8000^2} $$ follows a chi-squared distribution with $49$ degrees of freedom, regardless of the mean income. Now $$ S\ge5000\Leftrightarrow \frac{49S^2}{8000^2}\ge49\left(\frac{5}{8}\right)^2\ . $$ Therefore, the probability that the sample standard deviation is at least $\$5000$ is $\ 1-\chi^2_{49}\left(\frac{1225}{64}\right)\ $. Likewise, the probability that the sample standard deviation is no more than $\$1000$ is $\ \chi^2_{49}\left(\frac{49}{64}\right)\ $.