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

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:

Calculate the population mean

So, we have:

Calculate the standard deviation

So, we have:

Calculate the z-scores for x = 0.69 and 0.72

So, we have:

The probability is then represented as:

Using z table of probabilities, we have:

Hence, the required probability is (b) 0.2222

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b. 0.2222

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

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 , in which n is the number of items.

In this problem, we have that:

So

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

has a pvalue of 0.6477

X = 0.69

has a pvalue of 0.4255

0.6477 - 0.4255 =  0.2222

So the corect answer is:

b. 0.2222

-2x-4

step-by-step explanation: