Question

TV sets: According to the Nielsen Company, the mean number of TV sets in a U.S. household in 2013 was 2.24 . Assume the standard deviation is 1.2 . A sample of 95 households is drawn. Use the Cumulative Normal Distribution Table if needed.

What is the probability that the sample mean number of TV sets is greater than 2? Round your answer to four decimal places.

What is the probability that the sample mean number of TV sets is between 2.5 and 3? Round your answer to four decimal places.

Find the 80th percentile of the sample mean. Round your answer to two decimal places.

Would it be unusual for the sample mean to be less than 2? Round your answer to four decimal places.

It ▼(Choose one) unusual because the probability of the sample mean being less than 2 is .

Answer 1

Solution :

Given that,

mean = = 2.24

standard deviation = = 1.2

n = 95

=   = 2.24

= / n = 1.2 / 95 = 0.123

a) P( > 2) = 1 - P( < 2)

= 1 - P[( - ) / < (2 - 2.24) / 0.123]

= 1 - P(z < -1.95 )

Using z table,    

= 1 - 0.0256

= 0.9744

b) P(2.5 < < 3)  

= P[(2.5 - 2.24) /0.123 < ( - ) / < (3 - 2.24) / 0.123)]

= P(2.11 < Z < 6.18)

= P(Z < 6.18 ) - P(Z < 2.11)

Using z table,  

= 1 - 0.9826   

= 0.0174

c) Using standard normal table,

P(Z < z) = 80%

= P(Z < z ) = 0.80

= P(Z < 0.8416 ) = 0.80  

z = 0.8416

Using z-score formula  

= z * +   

= 0.8416 *0.123 + 2.24

= 2.34

d) P( < 2) = P(( - ) / < (2 - 2.24) / 0.123 )

= P(z < -1.95)

Using z table,

= 0.0256

Yes, it would be unusual because less than 5% of all such samples have means less than 2