Question

According to a report from Microsoft, 24% of PCs worldwide are not adequately protected by antivirus software.19 Suppose 200 PCs from around the world are selected at random.
a. Find the distribution of the sample proportion of PCs that
are not adequately protected.
b. Find the probability that the sample proportion is less
than 0.20.
c. Find the probability that the sample proportion is more
than 0.29.
d. Find a value v such that the probability the sample
proportion is less than v is 0.01.

Answer 1

Answer:

Given that:

(a)  Find the distribution of the sample proportion of PCs that are not adequately protected.

n = sample size = 200

p = sample proportion = 0.24

q = 1 - p = 0.76

SE =

The distribution of sample proportions is Normal Distribution with mean = 0.24 and SD = 0.0302.

(b) Find the probability that the sample proportion is less than 0.20.

To find P(p < 0.20):

Transforming to Standard Normal Variate:
Z = (0.20 - 0.24)/0.0302 = - 1.3245

Since z is negative, it lies on LHS of mid value.

Area Under Standard Normal Curve gives area from mid value to Z = 1.3245 on LHS of mid value as area = 0.4066

So,

P(p < 0.0.20) = 0.5 - 0.4066

= 0.0934

(c) Find the probability that the sample proportion is more than 0.29.

To find P(p>0.29):

Z = (0.29 - 0.24)/0.0302 = 1.6556

Since Z is positive, it lieson RHS of mid value. Table gives area from mid value to Z = 1.6556 on RHS of mid value as area = 0.4515.

So, P(p >0.29) = 0.5 - 0.4515

= 0.0485

(d) Find a value v such that the probability the sample proportion is less than v is 0.01.

To find \nu such that P(p < ) = 0.01.

Area from Z to left extreme = 0.01

i.e., area from mid value to Z is 0.5 - 0.01 = 0.49.

From Table, Z value corresponding to area = 0.49 is Z = 2.33.Since Z is on LHS of midvalue, it is negative.

So, Z = - 2.33