Question

Health care issues are receiving much attention in both academic and political arenas. A sociologist recently conducted a survey of citizens over 60 years of age whose net worth is too high to qualify for government health care but who have no private health insurance. The ages of 25 uninsured senior citizens were as follows: Suppose the mean and standard deviation are 74.0 and 9.7, respectively. If we assume that the distribution of ages is bell shaped, what percentage of the respondents will be between 64.3 and 93.4 years old

Answer 1

About 95% of the uninsured senior citizens in the survey who are over 60 years old are estimated to be between the ages of 64.3 and 93.4 based on the normal distribution with a mean of 74.0 and a standard deviation of 9.7, using the empirical rule.

Step-by-step explanation:

The question is asking to find the percentage of respondents aged between 64.3 and 93.4 years old based on a normal distribution with a mean of 74.0 and a standard deviation of 9.7. Using the 68-95-99.7 rule (empirical rule) for normal distributions, approximately 68% of data falls within one standard deviation from the mean, 95% within two standard deviations, and 99.7% within three standard deviations.

Since the interval from 64.3 to 93.4 years old is within approximately plus or minus two standard deviations away from the mean (64.3 = 74 - 9.7 and 93.4 = 74 + 9.7 * 2), we can estimate that about 95% of the respondents are likely within this age range.

Answer 2

P ( 64.3 ≤ X ≤ 93.4 ) = 0.8185 = 81.85 %

Step-by-step explanation:

To find the percentage up to 93.4 years old

P ( X ≤ 93.4 ) = ??

z (s) = ( x - μ ) / σ

z(s) = ( 93.4 - 74 ) / 9.7

z(s) = 19.4/ 9.7

z(s) = 2

Now with z (s) = 2 we find in the z-table the percentage 0,9772

P ( X ≤ 93.4 ) = 0.9772 = 97.72 %

Now to find the percentage above 64.3

P ( X ≤ 64.3 )

To find z₁(s) = ( 64.3 - 74 ) / 9.7

z₁(s) = - 9.7/9.7

z₁(s) = - 1

From z table P ( X ≤ 64.3 ) = 0.1587

Then as the part of the curve up to z = 2 ( area 0.9772 contains the area below z = -1 we need to subtract 0.9772 - 0.1587 )

Then the percentage between 64.3 and 93.4 is:

P ( 64.3 ≤ X ≤ 93.4 ) = 0.9772 - 0.1587 = 0.8185

P ( 64.3 ≤ X ≤ 93.4 ) = 0.8185 = 81.85 %