Question

According to a government study among adults in the 25- to 34-year age group, the mean amount spent per year on reading and entertainment is $1936. Assume that the distribution of the amounts spent follows the normal distribution with a standard deviation of $438. Refer to the table in Appendix B.1. (Round z-score computation to 2 decimal places and the final answer to 2 decimal places.)

a. What percentage of the adults spend more than $2180 per year on reading and entertainment?

Percentage            %

b. What percentage spend between $2180 and $3000 per year on reading and entertainment?

Percentage            %

c. What percentage spend less than $1000 per year on reading and entertainment?

Percentage            %

Answer 1

Solution :

Given that ,

mean = = 1936

standard deviation = = 438

a) P(x >2180 ) = 1 - p( x< 2180 )

=1- p P[(x - ) / < (2180 - 1936) / 438 ]

=1- P(z < 0.56 )

Using z table,

= 1 - 0.7123

= 0.2877

= 28.77%

b) P(2180 < x < 3000 ) = P[(2180 - 1936) / 438) < (x - ) /  < (3000 - 1936) / 438) ]

= P(0.56 < z < 2.43)

= P(z < 2.43 ) - P(z < 0.56)

Using z table,

= 0.9925 - 0.2877

= 0.7048

= 70.48%

c) P(x < 1000 ) = P[(x - ) / < (1000 - 1936) / 438]

= P(z < - 2.14)

Using z table,

= 0.0162

= 1.62%