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

Where the white dots are is where your points should be plotted.a translation is simply just moving something from one place to another. in this case, you would move each plot point of the original shape 2 units up, and 5 units to the left.

Answer from: Quest

The answer would be the second choice : )

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