Mathematics, 20.11.2019 02:31 agreisha

Suppose a simple random sample of size n=1000 is obtained from a population whose size is nequals=1,000,000 and whose population proportion with a specified characteristic is p equals 0.58 p=0.58. complete parts (a) through (c) below.

(a)-describe the sampling distribution of p.

(b)-what is the probability of obtaining x = 790 or more individuals with the characteristic?

(c)-what is the probability of obtaining x = 740 or fewer individuals with the characteristic?

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a) By the Central Limit Theorem, we have that the sampling distribution of p is approximately normal with mean $\mu = 580$ and standard deviation s = 15.61

b) There is a 33.36% probability of obtaining x = 790 or more individuals with the characteristic.

c) There is a 62.55% probability of obtaining x = 740 or fewer individuals with the characteristic,

Step-by-step explanation:

For each person of the sample, there are only two possible outcomes. Either they have the specified characteristic, or they do not. This means that we use the binomial probability distribution to solve this problem.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be 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.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X), \sigma = \sqrt{V(X)}$ .

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 $s = \frac{\sigma}{\sqrt{n}}$

(a)-Describe the sampling distribution of p.

First we use the data to find the expected value and standard deviation for the population.

Problem specifies that p = 0.58 .

Using the binomial approximation to the normal:

$\mu = E(X) = np = 1000000*0.58 = 580000$

The standard deviation of the population is

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \ssqrt{1000000*0.58*0.42} = 493.55$

For the sample

E(X) = 1000*0.58 = 580

$s = \frac{156.08}{\sqrt{1000}} = 15.61$

By the Central Limit Theorem, we have that the sampling distribution of p is approximately normal with mean $\mu = 580$ and standard deviation s = 15.61

(b)-What is the probability of obtaining x = 790 or more individuals with the characteristic?

This probability is 1 subtracted by the pvalue of Z when X = 790