Question

A manufacturer knows that their items have a lengths that are skewed right, with a mean of 16 inches, and standard deviation of 4.1 inches.If 43 items are chosen at random, what is the probability that their mean length is greater than 17 inches?(Round answer to four decimal places)

Answer 1

Answer: 0.492

Explanation:

H = 16

0 = 4.1

P(x > 17)=?

2 = -

0

17 -16

Z = -

4.1

1

Z =

4.1

z = 0.244

U sing the table for z

P(x > 17)=0.4927

So, the probability of the length is greater than 17 inches is 0.492

Answer 2

`\begin{gathered} \mu=16 \\ \sigma=4.1 \\ P(x>17)\text{=?} \\ z=(x-\mu)/(\sigma) \\ z=(17-16)/(4.1) \\ z=(1)/(4.1) \\ z=0.244 \\ U\sin g\text{ the table for z } \\ P(x>17)\text{=}0.4927 \\ So,\text{ the probability of the length is greater than 17 inches is 0.4927} \end{gathered}`