Answer:
4

5

Step-by-step explanation:
From the question we are told that
The percentage of hair dryers that are defective is p=2% 
The sample size is 
The random number is x = 219
The mean of this data set is evaluated as

substituting values


The standard deviation is evaluated as
![\sigma = \sqrt{[\mu (1 -p)]}](https://tex.z-dn.net/?f=%5Csigma%20%20%3D%20%20%5Csqrt%7B%5B%5Cmu%20%281%20-p%29%5D%7D)
substituting values
![\sigma = \sqrt{[200 (1 -0.02)]}](https://tex.z-dn.net/?f=%5Csigma%20%20%3D%20%20%5Csqrt%7B%5B200%20%281%20-0.02%29%5D%7D)

Since it is a one tail test the degree of freedom is
df = 0.5
So



Now applying normal approximation

substituting values


From the z-table

Considering Question 5
The random number is x = 90
The mean is 
Where n = 100
and p = 0.85
So

The standard deviation is evaluated as
![\sigma = \sqrt{[\mu (1 -p)]}](https://tex.z-dn.net/?f=%5Csigma%20%20%3D%20%20%5Csqrt%7B%5B%5Cmu%20%281%20-p%29%5D%7D)
substituting values
Since it is a one tail test the degree of freedom is
df = 0.5
Now applying normal approximation

substituting values


From the z-table
