Question

A manufacturer knows that their items have a lengths that are approximately normally distributed, with a mean of 5.8 inches, and standard deviation of 0.5 inches. If 31 items are chosen at random, what is the probability that their mean length is greater than 5.6 inches? (Round answer to four decimal places)

Answer 1

Answer: 0.9871

Step-by-step explanation:

Given : A manufacturer knows that their items have a lengths that are approximately normally distributed with

$\mu=5.8 \text{ inches}$

$\sigma=0.5\text{ inches}$

Sample size : $n=31$

Let x be the length of randomly selected item.

z-score : $z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}$

$z=\dfrac{5.6-5.8}{\dfrac{0.5}{\sqrt{31}}}\approx-2.23$

The probability that their mean length is greater than 5.6 inches by using the standard normal distribution table

= $P(x>5.6)=P(z>-2.23)=1-P(z$

$=1-0.0128737=0.9871263\approx0.9871$

Hence, the probability that their mean length is greater than 5.6 inches is 0.9871.