Mathematics, 01.04.2020 20:30 jillericamurph46

Suppose a batch of metal shafts produced in a manufacturing company have a standard deviation of 1.9 and a mean diameter of 200 inches. If 78 shafts are sampled at random from the batch, what is the probability that the mean diameter of the sample shafts would differ from the population mean by less than .2 inches?

Answers: 1

Show answers

Answers

64.76% probability that the mean diameter of the sample shafts would differ from the population mean by less than .2 inches

Step-by-step explanation:

To solve this question, we need 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 normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

$\mu = 200, \sigma = 1.9, n = 78, s = \frac{1.9}{\sqrt{78}} = 0.2151$

What is the probability that the mean diameter of the sample shafts would differ from the population mean by less than .2 inches?

This is the pvalue of Z when X = 200 + 0.2 = 200.2 subtracted by the pvalue of Z when X = 200 - 0.2 = 199.8. So

X = 200.2

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

By the Central Limit Theorem

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

$Z = \frac{200.2 - 200}{0.2151}$

Z = 0.93

Z = 0.93 has a pvalue of 0.8238

X = 199.8

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

$Z = \frac{199.8 - 200}{0.2151}$

Z = -0.93

Z = -0.93 has a pvalue of 0.1762

0.8238 - 0.1762 = 0.6476

64.76% probability that the mean diameter of the sample shafts would differ from the population mean by less than .2 inches

Answer from: Quest

u rich

step-by-step explanation:

Answer from: Quest

26?

step-by-step explanation:

Another question on Mathematics

Mathematics, 21.06.2019 17:30

If f(x)=3x+1 and f^-1=x-1/3, then f^-1(7)=

Answers: 1

Answer

Mathematics, 21.06.2019 17:30

Monthly water bills for a city have a mean of $108.43 and a standard deviation of $32.09. find the probability that a randomly selected bill will have an amount greater than $155, which the city believes might indicate that someone is wasting water. would a bill that size be considered unusual?

Answers: 2

Answer

Mathematics, 21.06.2019 18:00

Janie has $3. she earns $1.20 for each chore she does and can do fractions of chores. she wants to earn enough money to buy a cd for $13.50. write an inequality to determine the number of chores, c, janie could do to have enough money to buy the cd.

Answers: 1

Answer

Mathematics, 21.06.2019 23:30

Petes dog weighed 30 pounds it then lost 16% of it's weight how much did pete lose

Answers: 2

Answer

You know the right answer?

Suppose a batch of metal shafts produced in a manufacturing company have a standard deviation of 1.9...

Questions