I was given this question in my probability homework.
Given a normal population whose
mean is $355$
standard deviation is $42$
Find the probability that a random sample of $5$ has a mean between $362$ and $372$
According to the formula I am using:
$$P(Z >\frac{362-355}{\frac{42}{\sqrt{5}}})$$
$$P(Z >\frac{372-355}{\frac{42}{\sqrt{5}}})$$
From these calculations I get:
$$0.37-0.91$$
now I looked up these values in my z-table, which yields:
$$0.3557-0.1814$$
The result is $$0.1743$$
I have this as my answer but the system won't accept it as a solution. So I don't know where the mistake is. I had someone else do the same problem nd we both got the same answer.
If observations are drawn from a population with mean $\mu = 355$ and standard deviation $\sigma = 42$, then a sample of size $n = 5$ will have a sample mean $\bar X$ that is normally distributed with mean $\mu = 355$ and standard deviation $\sigma/\sqrt{n} = 42/\sqrt{5}$. Then the probability that the sample mean is between $362$ and $372$ is $$\begin{align*} \Pr[362 < \bar X < 372] &= \Pr\left[\frac{362 - 355}{42/\sqrt{5}} < \frac{\bar X - \mu}{\sigma/\sqrt{n}} < \frac{372 - 355}{42/\sqrt{5}}\right] \\ &= \Pr\left[ \frac{\sqrt{5}}{6} < Z < \frac{17\sqrt{5}}{42} \right] \\ &\approx \Phi(0.905075) - \Phi(0.372678) \\ &\approx 0.171981. \end{align*}$$ It would seem that your answer is not sufficiently precise due to rounding at too early a stage in your calculations.