Question

Suppose that the mean caloric intake for Americans is 2,500 calories per day, with a standard devia-
tion of 175 calories per day. Assume that caloric intakes for Americans are approximately normally
distributed.
(a) (3 points) Find the probability that a randomly selected American consumes more than 2,250
calories per day.
(b) (3 points) Find the probability that a randomly selected American consumes at most 1,950 calo-
ries per day.
(c) (3 points) Find the probability that a randomly selected American consumes between 2,550 calo-
ries per day and 3,050 calories per day.

Answer 1

Solution :

mean = = 2500

standard deviation = =175

a) P(x > 2250) = 1 - p( x< 2250 )

=1- p [(x - ) / < (2250 -2500) /175 ]

=1- P(z < -1.43)

= 1 - 0.0764 = 0.9236

probability = 0.9236

b)

P(x 1950)

= P[(x - ) / (1950-2500) /175 ]

= P(z -3.43)

=0.0003

probability = 0.0003

c)

P( 2550 < x < 3050 ) = P[(2550 -2500)/175 ) < (x - ) /  < (3050 - 2500) /175 ) ]

= P( 0.29 < z < 3.14 )

= P(z <3.14 ) - P(z < 0.29 )

Using standard normal table

= 0.9992 - 0.6141 = 0.3851

Probability = 0.3851