A sociologist recently conducted a survey of citizens over 60 years of age who have net worths too high to qualify for Medicaid but have no private health insurance. The ages of the 25 uninsured senior citizens were as follows: Suppose the mean and standard deviation are 74.04 and 9.75, respectively. If we assume that the distribution of ages is mound-shaped and symmetric, what percentage of the respondents will be between 64.29 and 93.54 years old

81.85% of the respondents will be between 64.29 and 93.54 years old

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be 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.

Mean and standard deviation are 74.04 and 9.75

This means that

What percentage of the respondents will be between 64.29 and 93.54 years old?

The proportion is the pvalue of Z when X = 93.54 subtracted by the pvalue of Z when X = 64.29. So

X = 93.54

has a pvalue of 0.9772.

X = 64.29

has a pvalue of 0.1587.

0.9772 - 0.1587 = 0.8185

0.8185*100 = 81.85%

81.85% of the respondents will be between 64.29 and 93.54 years old

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step-by-step explanation: