Question

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?

Answer 1

Answer:

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:

$n = 1000, p = 0.76$

So

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

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

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