Mathematics, 01.07.2020 15:01 Lydiaxqueen

Suppose a simple random sample of size n = 1000 is obtained from a population whose size i: N = 2,000,000 and whose population proportion with a specified characteristic is p = 0.76. Complete parts (a) through (c) below. 1. What is the probability of obtaining x = 790 or more individuals with the characteristic?
2. What is the probability of obtaining x =740 or fewer individuals with the characteristic?

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Answer from: sonnie28

1. 1.46% probability of obtaining x = 790 or more individuals with the characteristic

2. 7.49% probability of obtaining x =740 or fewer individuals with the characteristic

Step-by-step explanation:

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 $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

In this problem, we have that:

n = 1000, p = 0.76

$\mu = E(X) = np = 1000*0.76 = 760$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{1000*0.76*0.24} = 13.51$

1. What is the probability of obtaining x = 790 or more individuals with the characteristic?

Using continuity correction, this is $P(X \geq 790 - 0.5) = P(X \geq 789.5)$ , which is 1 subtracted by the pvalue of Z when X = 789.5. So

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

$Z = \frac{789.5 - 760}{13.51}$

Z = 2.18

Z = 2.18 has a pvalue of 0.9854

1 - 0.9854 = 0.0146

1.46% probability of obtaining x = 790 or more individuals with the characteristic.

2. What is the probability of obtaining x =740 or fewer individuals with the characteristic?

Using continuity correction, this is $P(X \leq 740 + 0.5) = P(X \leq 740.5)$ , which is the pvalue of Z when X = 740.5. So

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

$Z = \frac{740.5 - 760}{13.51}$

Z = -1.44

Z = -1.44 has a pvalue of 0.0749

7.49% probability of obtaining x =740 or fewer individuals with the characteristic

Answer from: Quest

Idon’t understand what you’re asking..

Answer from: Quest

what are ur choices?

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