Mathematics, 06.03.2020 23:56 ziazozaya

A manufacturer knows that their items have a normally distributed length, with a mean of 13.1 inches, and standard deviation of 4.1 inches. If 25 items are chosen at random, what is the probability that their mean length is less than 11.1 inches

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Answer from: failingschooll0l

0.73% probability that their mean length is less than 11.1 inches

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation, which is also called standard error $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 13.1, \sigma = 4.1, n = 25, s = \frac{4.1}{\sqrt{25}} = 0.82$

What is the probability that their mean length is less than 11.1 inches

This is the pvalue of Z when X = 11.1. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{11.1 - 13.1}{0.82}$

Z = -2.44

Z = -2.44 has a pvalue of 0.0073.

0.73% probability that their mean length is less than 11.1 inches

Answer from: Quest

tickets for the harlem globetrotters

show at michigan state university in

2010 cost $16, $23, or, for vip seats, $40.

total revenue = 16x + 23y + 40z

if 9 times as many $16 tickets were sold

as vip tickets,

x = 9z => x – 9z = 0, if 9 times as many $16 tickets were sold

as vip tickets,

x = 9z => x – 9z = 0, sales of all three kinds of tickets would

total $46,575.

16x + 23y + 40z = 46575, how many of each kind of ticket would

have been sold?

solve for x, y and z

x − 9z = 0

x − y − z = 55

16x + 23y + 40z = 46575

x = 1170

y = 985

z = 130

welcome!

Answer from: Quest

3 1/3 x 2 3/5 = 8 2/3

step-by-step explanation:

just multiply the first number by the second number to get your answer.

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A manufacturer knows that their items have a normally distributed length, with a mean of 13.1 inches...