Mathematics, 20.02.2020 19:37 rujirasri3620

4. Household size in the United States has a mean of 2.6 people and standard deviation of 1.4 people. However this distribution is skewed right as the smallest possible value is a household of 1 person butthe largest households can be very large indeed.(a) Suppose one household in the US is to be randomly selected. Can you calculate the probabilitythat the household size is higher than 4? If yes, calculate it; if no, explain why not.(b) Suppose 120 households are randomly selected and the average values in the sample is calculated. What is the expected value of mean size in the sample of 120 households?(c) Suppose 120 households are randomly selected and the average values in the sample is calculated. What is the standard deviation of mean size in the sample of 120 households?(d) What is the probability that the mean size of a random sample of 120 households is more than 3?

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Answer from: angie07253

a) Cannot calculate, because the population is not normal.

b) 2.6 people

c) 0.1278 people

d) 0.9% probability that the mean size of a random sample of 120 households is more than 3.

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 random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 2.6, \sigma = 1.4$

(a) Suppose one household in the US is to be randomly selected. Can you calculate the probabilitythat the household size is higher than 4? If yes, calculate it; if no, explain why not.

The population is not normal, so we cannot use the z-score formula to find this probability. So we cannot calculate this probability.

(b) Suppose 120 households are randomly selected and the average values in the sample is calculated.What is the expected value of mean size in the sample of 120 households?

By the Central Limit Theorem, we have that the mean is the same as the mean of the population, that is, 2.6 people.

(c) Suppose 120 households are randomly selected and the average values in the sample is calculated.What is the standard deviation of mean size in the sample of 120 households?

By the Central Limit Theorem

$s = \frac{1.4}{\sqrt{120}} = 0.1278$

(d) What is the probability that the mean size of a random sample of 120 households is more than 3?

This is 1 subtracted by the pvalue of Z when X = 3. So

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

By the Central Limit Theorem

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

$Z = \frac{3 - 2.6}{0.1278}$

Z = 3.13

Z = 3.13 has a pvalue of 0.9991.

1 - 0.9991 = 0.0009

0.9% probability that the mean size of a random sample of 120 households is more than 3.

Answer from: Quest

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Answer from: Quest

investment b is an ordinary annuity with $3,909.64 in the account after one year.

step-by-step explanation:

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