Question

The average amount spent by a family of two on food per month is $500 with a standard deviation of $75. Assuming that the food costs are normally distributed,

what is the probability that a family spends less than $410 per month?

what is the probability that a family spends great than $610 per month?

what is the probability that a family spends $500±50(between $450 and $550) per month?

Answer 1

Solution :

Given that ,

mean = = 500

standard deviation = = 75

P(x < 410 )

= P[(x - ) / < ( 410 - 500 ) / 75 ]

= P(z < -1.2)

Using z table,

= 0.1151

probability = 0.1151

b.

P(x > 610 ) = 1 - P( x < 610 )

=1- P[(x - ) / < ( 610 - 500) / 75 ]

=1- P(z < 1.47 )

Using z table,

= 1 - 0.9292

= 0.0708

probability = 0.0708

c.

P( 450 < x < 550 )

= P[( 450 - 500) / 75 ) < (x - ) /  < ( 550 - 500) / 75 ) ]

= P( -0.67 < z < 0.67)

= P(z < 0.67 ) - P(z < -0.67 )

Using z table,

= 0.7486 - 0.2514

= 0.4972

probability = 0.4972