Question

According to past data, the 15% of all college students in California are business majors. Suppose a random sample of 200 California college students is taken.

a) What information about this sample allows us to use the normal distribution for our sampling distribution?

b) Calculate the standard error. Round to two places for ease.

c) What is the probability that the sample of 200 gives a sample proportion of 18% or higher? Show your calculator function and entries. Round to 4 places.

Answer 1

a) We use the normal distribution for our sampling distribution if np 10 and n(1-p) 10

Here , n = 200 , p = 15% = 0.15

np= 200*0.15 = 30 (   10 ) and n(1-p) = 200*0.85 = 170 (   10 )   

Condition satisfied. So we can use Normal distribution for our sampling distribution.

Sampling distribution of sample proportion if approximately normal with mean and

standard deviation =

b )

Standard Error ( SE ) OR Standard deviation :

c)

We have , mean   = 0.15

standard deviation = = 0.02525

We have to find P( >= 0.18 )

P( >= 0.18 ) = 1 - P( < 0.18 )

Using Excel function, =NORMSDIST( x, mean , standard deviation , 1 )

P( < 0.18 ) = NORMDIST( 0.18, 0.15, 0.02525,1) = 0.88261

So,  P( >= 0.18 ) = 1 - 0.88261 = 0.1174

The probability that the sample of 200 gives a sample proportion of 18% or higher is 0.1174