a manufacturer knows that their items have a lengths that are skewed right, with a mean of 18.8 inches, and standard deviation of 6.2 inches. if 48 items are chosen at random, what is the probability that their mean length is greater than 20.1 inches?



Answer :

The probability that their mean length is greater than 20.1 inches is 0.32.

Standard deviation:

The standard deviation is a measure of the amount of variation or dispersion of a set of values. A low standard deviation indicates that the values tend to be close to the mean of the set, while a high standard deviation indicates that the values are spread out over a wider range.

The given value are:

Mean = 18.8 inches

Standard deviation = 6.2 inches

We have to find the probability that their mean length is greater than 20.1 inches.

Variance of sample mean = 6.2 / 48

                                           = 0.13

S.D of sample mean=   [tex]\sqrt{0.13}[/tex]

                                 = 0.36

Z score for the length of 20.1 is ( 20.1 - 18.8) / 0.36 = 0.468

The p-value for 0.468 in the standard normal table is 0.32.

Therefore 0.32 chance of the sample means is greater than 20.1 inches.

To know more about the standard deviation refer to the link given below:

https://brainly.com/question/475676

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