Question

(URGENT )The average number of chips in a large bag is 41 chips with a standard deviation of 3 chips. You sample 30 bags of chips. Assuming a normal distribution, what is the probability the sample mean is 40 chips?

Answer 1

Let mean be $\mu _x$ and sample mean be $\overline{X}$.

Given, $\mu_x = 41$ and $\sigma = 3$,

$n = 30$ (number of samples)

to find: $P( \overline{X} = 40)$

If the standard deviation, σ, is known we can transform to a standard normal variable,

$\overline{X } \sim N(\mu_x, \frac{\sigma}{\sqrt{n}})$

$\overline{X } \sim N(41, 0.5477)$

Therefore,

$P( \overline{X} = 40) = \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{(x - \mu)^2}{2 \sigma^2}} \\ \\ = \frac{1}{0.5477\sqrt{2\pi}} e^{-\frac{(40 - 41)^2}{2 *0.5477^2}} \\ \\ = \frac{1}{1.373}e^{-1.667} \\ \\ = 0.138$