Question

Suppose that the time spent by children in front of the television set per year has a distribution   N(1500, 250); i.e, normal with mean 1500 and standard deviation 250.

What proportion of children spend less than 1000 hours in front of the television? (Express your answer as a decimal, do not convert to percent. Keep four decimal places.)

What proportion of children spend more than 1700 hours in front of the television? (Express your answer as a decimal, do not convert to percent. Keep four decimal places.)

What proportion of children spend between 1500 and 1700 hours in front of the television? (Express your answer as a decimal, do not convert to percent. Keep four decimal places.)

Answer 1

Solution :

Given that ,

mean = = 1500

standard deviation = = 250

a) P(x < 1000) = P[(x - ) / < (1000 - 1500) / 250 ]

= P(z < -2.00)

Using z table,

= 0.0228

b) P(x > 1700) = 1 - p( x< 1700)

=1- p P[(x - ) / < (1700 - 1500) / 250 ]

=1- P(z < 0.80 )

Using z table,

= 1 - 0.7881

= 0.2119

c) P(1500 < x < 1700) = P[(1500 - 1500)/ 250 ) < (x - ) /  < (1700 - 1500) / 250) ]

= P( 0 < z < 0.80)

= P(z < 0.80) - P(z < 0)

Using z table,

= 0.7881 - 0.5

= 0.2881