Question

In a developing country, 19% of the entire population has high
speed access to the internet. Random samples of size 200 are selected from the country’s
population in order to estimate the proportion of the population with high-speed internet
access.
a) What is the expected sample proportion? (Do not use the CLT.)
b) What is the standard deviation of the sample proportion? (Do not use the CLT.)
c) What is the probability that the sample proportion is (strictly) less than 14%? (Do not
use the CLT.)
d) Use the Central Limit Theorem (CLT) to estimate the probability that the sample
proportion is between 9% and 29%?
e) Use the CLT to estimate the value p such that the probability that the sample propor
tion is above p is 90%.

Note : Please explain C in details

Answer 1

Answer to part a)

n = 200

p = 19% = 0.19

expected proportion = n * p

expected proportion = 200*0.19 = 38

.

Part b)
Standard deviation = sqrt(p * (1-p)/ n)
Standard deviation = sqrt(0.19*(1-0.19)/200)

[sqrt = square root]

Standard deviation = 0.02774

.

Part c)

z = (p^ - p) / Standard deviation

p^ = 0.14

p = 0.19

SD = 0.02774

z = (0.14-0.19)/0.02774

z = -1.80

.

refer to Z table , p(z < -1.80) = 0.0359

.

Part d)

P(0.09 < p < 0.29) = P(p < 0.29) - P(p < 0.09)

z1 = (0.09-0.19)/0.02774 = -3.60

z2 = (0.29-0.19)/0.02774 = +3.60

P(z< -3.60) = 0.00016

P(z < +3.60) = 0.99984

P(0.09 < p < 0.29) = 0.99984-0.00016 = 0.99968

.

Part e)

probability that sample proportion is above p is 90%

or in orders that the sample proportion is less than p is 10%

Z value corresponding to probability 10% is -1.28

.

Formula of Z = ( x - M ) / SD

-1.28 = (x - 38) / 0.02774

-0.0355 = x - 38

x = 38-0.0355

x = 37.9645