Question

In: Math

1. Suppose a random sample of 100 Utah families yields a total fertility rate (the average...

1. Suppose a random sample of 100 Utah families yields a total fertility rate (the average number of children born per woman) of 2.71. Given what the Central Limit Theorem tells us about the relationship between sample statistics and population parameters, we can assume the total fertility rate for all Utah families is

Exactly 2.71. No more, no less.

Less than 2.71, as the sampling distribution for the TFR is always negatively skewed

likely close to 2.71, though not necessarily EXACTLY 2.71

Far too high relative to the number of students our public education system can realistically support (this is NOT the correct answer)

2. Suppose you're interested in tracking Americans' church attendance. A sample of 900 Americans finds that 22 percent of respondents report attending church at least once per week. If this poll has a margin of error of plus/minus three percent, we can be fairly certain that the percentage of all Americans who attend church at least once per week falls between _______ percent and _______ percent.

Group of answer choices

19; 25

21.7; 22.3

16; 28

There's simply no way of knowing because everyone lies about how often they go to church (this is not the correct answer, although it's probably true).

3. Suppose weights of high school wrestlers are normally distributed with a mean of 145 pounds and a standard deviation of 15 pounds. Approximately what percent of wrestlers weigh between 135 and 150 pounds?

Group of answer choices

37.81 percent

74.75 percent

25.25 percent

0.3781 percent

4. Suppose weights of high school wrestlers are normally distributed with a mean of 145 pounds and a standard deviation of 15 pounds. What weight (approximately) corresponds with the 75th percentile?

Group of answer choices

134.883

140.22

149.78

155.117

Solutions

Expert Solution

1 -

Central limit theorem states that - for a relatively large sample size (more than 30), the sample mean is an unbiased estimator of the population mean.

In that case, we can assume the population mean to be exactly 2.71. No more, no less

2 -

Given the sample is random sample and there is no response bias from the sample. We can consider the estimate to be an unbiased estimate and therefore, the confidence interval estimate becomes -

point estimate +/- margin of error

= 19 +/- 3

= 16 to 25

3 -

mean = 145

std dev = 15

Proportion of wrestlers between 135 and 150 pounds -

P (Z between ((145 - 135) / 15 and (145 - 150) / 15)

= P (-0.33 < Z < 0.66)

= P(Z < 0.66) - P(Z > -0.33)

= 0.75 - 0.38 approximately

= 0.37 approximately

= 37.81 percent

4 -

z value for 75th percentile = z value for p value of 0.75

= 0.69

(X - mean) / std dev = 0.69

=> (X - 145) / 15 = 0.69

=> X = 155.3

Weight of 155.3 pounds corresponds the 75th percentile


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