Mathematics, 24.06.2021 20:50 hurricane01

It has been estimated that 17.0 % of mutual fund shareholders are retired persons. Assuming a simple random sample of 400 shareholders has been selected: a. What is the probability that at least 20 % of those in the sample will be retired? b. What is the probability that between 15 % and 21 % of those in the sample will be retired?

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Answer from: bradydodson47

a) 0.0548 = 5.48% probability that at least 20 % of those in the sample will be retired.

b) 0.8388 = 83.88% probability that between 15 % and 21 % of those in the sample will be retired.

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 $s = \sqrt{\frac{p(1-p)}{n}}$

17.0 % of mutual fund shareholders are retired persons.

This means that p = 0.17

Assuming a simple random sample of 400 shareholders has been selected

This means that n = 400

Mean and standard deviation:

$\mu = p = 0.17$

$s = \sqrt{\frac{0.17*0.83}{400}} = 0.0188$

a. What is the probability that at least 20 % of those in the sample will be retired?

This is the 1 p-value of Z when X = 0.2. So

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

By the Central Limit Theorem

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

$Z = \frac{0.2 - 0.17}{0.0188}$

Z = 1.6

Z = 1.6 has a p-value of 0.9452

1 - 0.9452 = 0.0548

0.0548 = 5.48% probability that at least 20 % of those in the sample will be retired.

b. What is the probability that between 15 % and 21 % of those in the sample will be retired?

This is the p-value of Z when X = 0.21 subtracted by the p-value of Z when X = 0.15.

X = 0.21

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

$Z = \frac{0.21 - 0.17}{0.0188}$

Z = 2.13

Z = 2.13 has a p-value of 0.9834

X = 0.15

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

$Z = \frac{0.15 - 0.17}{0.0188}$

Z = -1.06

Z = -1.06 has a p-value of 0.1446

0.9834 - 0.1446 = 0.8388

0.8388 = 83.88% probability that between 15 % and 21 % of those in the sample will be retired.

Answer from: Quest

i think this could be a possible answer (1 / 84) -1

good luck! : d

Answer from: Quest

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If f(x)= 3x^2and g(x) = x+2, find (f•g)(x)

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