Question

The population mean length of piece of yarn is 230 inches with standard deviation of 14 inches. A sample of 60 pieces of the yarn was taken. What is the probability that the sample mean differs from the population mean by atleast 1.25 inches?

Answer 1

Let $X=$ the sample mean: $X\sim N(\mu, \sigma^2/n).$ Then $Z=\dfrac{X-\mu}{\sigma/\sqrt{n}}\sim N(0, 1)$

Given $\mu=230\ inches, \sigma=14\ inches, n=60$

$$P(X\leq\mu-1.25)=P(Z\leq\dfrac{\mu-1.25-\mu}{\sigma/\sqrt{n}})$$

$$=P(Z\leq\dfrac{-1.25}{14/\sqrt{60}})\approx P(Z\leq-0.691604)$$

$$\approx0.2445930$$

$$P(X\geq\mu+1.25)=1-P(X<\mu+1.25)$$

$$=1-P(Z<\dfrac{\mu+1.25-\mu}{\sigma/\sqrt{n}})$$

$$=1-P(Z<\dfrac{1.25}{14/\sqrt{60}})\approx1- P(Z\leq0.691604)$$

$$\approx0.2445930$$

$P(X\leq230-1.25\ or\ X\geq 230+1.25)=2\cdot0.2445930=0.489186$