Suppose it is known that 8 out of the 20 teams in the sample had a season winning percentage better than 0.500.
(a) find a 95% confidence interval for the true proportion of all teams that had a season winning percentage better than 0.500.

The 95% confidence interval for the true proportion of all teams that had a season winning percentage better than 0.500 is (0.1853, 0.6147).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of , and a confidence interval , we have the following confidence interval of proportions.

In which

Z is the zscore that has a pvalue of .

For this problem, we have that:

8 out of the 20 teams in the sample had a season winning percentage better than 0.500. This means that .

95% confidence interval

So , z is the value of Z that has a pvalue of , so .

The lower limit of this interval is:

The upper limit of this interval is:

The 95% confidence interval for the true proportion of all teams that had a season winning percentage better than 0.500 is (0.1853, 0.6147).

hello,

b) 62.47% of the students are between 14 and 18 years old.

step-by-step explanation:

mean = 15 and standard deviation = 2

student between 14 to 18

formula used z-score

for age 14 , z score

using z-table , p(z> -0.5)=0.3085

for age 18 , z score

using z-table , p(z< 1.5)=0.9332

p(-0.5< z< 1.5)=0.9332-0.3085 = 0.6247

thus, 62.47% of the students are between 14 and 18 years old.

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