Answer:
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
