Mathematics, 01.07.2021 18:20 xDANTEx2456

The manager of a donut store believes that 35% of the customers are first-time customers. A random sample of 150 customers will be used to estimate the proportion of first-time customers. Assuming this belief is correct, what is the probability that the sample proportion will be between 0.2 and 0.4

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Answer from: annette211pdd8v9

0.8996 = 89.96% probability that the sample proportion will be between 0.2 and 0.4

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 normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes 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.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

The manager of a donut store believes that 35% of the customers are first-time customers.

This means that p = 0.35

Sample of 150 customers

This means that n = 150

Mean and standard deviation:

$\mu = p = 0.35$

$s = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.35*0.65}{150}} = 0.0389$

What is the probability that the sample proportion will be between 0.2 and 0.4?

p-value of Z when X = 0.4 subtracted by the p-value of Z when X = 0.2.

X = 0.4

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

By the Central Limit Theorem

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

$Z = \frac{0.4 - 0.35}{0.0389}$

Z = 1.28

Z = 1.28 has a p-value of 0.8997

X = 0.2

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

$Z = \frac{0.2 - 0.35}{0.0389}$

Z = -3.85

Z = -3.85 has a p-value of 0.0001

0.8997 - 0.0001 = 0.8996

0.8996 = 89.96% probability that the sample proportion will be between 0.2 and 0.4

Answer from: Quest

people grow faster than other people.

step-by-step explanation:

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Is it normal to have this much hair at age 14 i am a boy

Answer from: Quest

Bi’m pretty sure hope it works

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