Question

In a local university, 70% of the students live in the dormitories. A random sample of 75 students is selected for a particular study. The probability that the sample proportion (the proportion living in the dormitories) is between 0.69 and 0.72 is a. 0.7778 b. 0.2222 c. 0.2347 d. 0.7653

Answer 1

Answer:

b. 0.2222

Step-by-step explanation:

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.

For a proportion p, we use $\mu = p, \sigma = \sqrt{\frac{p(1-p)}{n}}$, in which n is the number of items.

In this problem, we have that:

$p = 0.7, n = 75$

So

$\mu = 0.7, \sigma = \sqrt{\frac{0.7*0.3}{75}} = 0.0529$

Between 0.69 and 0.72

This is the pvalue of Z when X = 0.72 subtracted by the pvalue of Z when X = 0.69. So

X = 0.72

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

$Z = \frac{0.72 - 0.7}{0.0529}$

$Z = 0.778$

$Z = 0.377$ has a pvalue of 0.6477

X = 0.69

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

$Z = \frac{0.69 - 0.7}{0.0529}$

$Z = -0.188$

$Z = -0.188$ has a pvalue of 0.4255

0.6477 - 0.4255 =  0.2222

So the corect answer is:

b. 0.2222

Answer 2

The required probability that the sample proportion (the proportion living in the dormitories) is between 0.69 and 0.72 is (b) 0.2222

The given parameters are:

$\mathbf{n = 75}$

$\mathbf{p = 70\%}$

Calculate the population mean

$\mathbf{\mu = p}$

So, we have:

$\mathbf{\mu = 70\%}$

$\mathbf{\mu = 0.7}$

Calculate the standard deviation

$\mathbf{\sigma = \sqrt{\frac{p(1 - p)}{n}}}$

So, we have:

$\mathbf{\sigma = \sqrt{\frac{70\% \times (1 - 70\%)}{75}}}$

$\mathbf{\sigma = \sqrt{\frac{0.7 \times 0.3}{75}}}$

$\mathbf{\sigma = \sqrt{\frac{0.21}{75}}}$

$\mathbf{\sigma = 0.053}$

Calculate the z-scores for x = 0.69 and 0.72

$\mathbf{z = \frac{x - \mu}{\sigma}}$

So, we have:

$\mathbf{z = \frac{0.69 - 0.7}{0.053} =-0.1887 }$

$\mathbf{z = \frac{0.72 - 0.7}{0.053} =0.3774}$

The probability is then represented as:

$\mathbf{P(0.69 < x < 0.72) = P(z<0.3774) - P(-0.1887)}$

Using z table of probabilities, we have:

$\mathbf{P(0.69 < x < 0.72) = 0.6477 - 0.4255 }$

$\mathbf{P(0.69 < x < 0.72) = 0.2222}$

Hence, the required probability is (b) 0.2222

Read more about probabilities at:

https://brainly.com/question/11234923