A university dean is interested in determining the proportion of students who receive some sort of financial aid. Rather than examine the records for all students, the dean randomly selects 200 students and finds that 118 of them are receiving financial aid. Use a 99% confidence interval to estimate the true proportion of students on financial aid, with limits rounded to four decimal places.

Step-by-step explanation:

Mean = np

Where

n = number of students

p = probability of success

n = 200

p = 118/200 = 0.59

mean = 200 × 0.59 = 118

Standard deviation, s = √npq

q = 1 - p = 1 - 0.59

q = 0.41

s = √(200 × 0.59 × 0.41)

s = √48.38 = 6.956

For a confidence level of 99%, the corresponding z value is 2.58. This is determined from the normal distribution table.

We will apply the formula

Confidence interval

= mean ± z ×standard deviation/√n

It becomes

118 ± 2.58 × 6.956/√200

= 118 ± 2.58 × 0.492

= 118 ± 1.2694

The lower end of the confidence interval is 118 - 1.2694 = 116.7306

The upper end of the confidence interval is 118 + 1.2694 = 119.2694

answer: huh

step-by-step explanation:

child does not get an education so she can use the molla on junk