A group of 40 bowlers showed that their average score was 192. Assume the population standard deviation is 8. Find the 95% confidence interval of the mean score of all bowlers.
A)(189.5, 194.5)
B)(186.5, 197.5)
C)(188.5, 195.6)
D)(187.3, 196.1)

Question

Quizplus · Accepted Answer

We can use the formula for a confidence interval:
CI = X ± (Z α/2)(σ/√n)
where X is the sample mean (192), Z α/2 is the z-score associated with a 95% confidence level (1.96), σ is the population standard deviation (8), and n is the sample size (40).

Plugging in the values, we get:
CI = 192 ± (1.96)(8/√40)
CI = 192 ± 2.95

So the 95% confidence interval for the true mean score of all bowlers is:
CI = (192 - 2.95, 192 + 2.95)
CI = (189.05, 194.95)

Rounded to one decimal place, we get (189.5, 194.5). Therefore, the best choice is A. We can use the formula for confidence interval of the mean:

CI = x̄ ± z*(σ/√n)

where x̄ is the sample mean, σ is the population standard deviation, n is the sample size, and z depends on the level of confidence desired.

For a 95% confidence interval, we can look up the z-value in a standard normal distribution table, which gives us z = 1.96.

Plugging in the given values, we get:

CI = 192 ± 1.96*(8/√40)

Simplifying, we get:

CI = 192 ± 2.81

So the confidence interval is (189.19, 194.81). Rounding to the nearest tenth, we get (189.5, 194.5).

Therefore, the best choice is A.

A Group of 40 Bowlers Showed That Their Average Score