Ms Kelly puts more than 5 apples in each bag. Write an inequality to represent the number of apples in a bag

It has been estimated that 17.0 % of mutual fund shareholders are retired persons. Assuming a simple random sample of 400 shareh

kkurt [141]

Answer:

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.

3 0

3 years ago

Hey can you please help me posted picture of question

horsena [70]

The answer for the exercise shown in the image attached is the third option, the option C, which is: C. 1/9
The explanation is shown below:
The information given in the problem above, says that the list of the numbers are randomly generated. Then, you have that there are nine numbers on the list, including the number zero, therefore, the probability that each number occurs is 1/9, so the correct answer is the option C.

4 0

3 years ago

A store randomly samples 602 shoppers over the course of a year and finds that 144 of them made their visit because of a coupon

suter [353]

Answer:

well as a percent its 23.92

Step-by-step explanation:

8 0

3 years ago

What is 0.16666666666 as a fraction

rosijanka [135]

Hello there.

What is 0.16666666666 as a fraction

Answer: 1/6