Question

A recent report in Pasadena Times indicated a typical family of four spends $490 per month on food. Assume the distribution of food expenditures for a family follows the normal distribution, with a standard deviation of $90 per month.

1. What percent of the families spend between $300 and $490 per month on food?
2. What is the probability that a families selected spends less than $430 per month on food?
3. What percent spend between $430 and $600 per month on food?
4. What is the probability of selecting a family that spends between $400 and $600 per month on food?

Answer 1

Solution :

Given that ,

mean = = 490

standard deviation = = 90

P(300 < x < 490) = P[(300 - 490)/ 90) < (x - ) /  < (490 - 490) /90 ) ]

= P(-2.11 < z < 0)

= P(z < 0 ) - P(z < -2.11 )

Using z table,

= 0.5 - 0.0174

= 0.4826

= 48.26%

P(x < 430 ) = P[(x - ) / < ( 430 - 490) / 90 ]

= P(z < -0.67)

Using z table,

= 0.2514

probability = 0.2514

P(430 < x < 600) = P[(430 - 490)/ 90 ) < (x - ) /  < (600 - 490) / 90) ]

= P(-0.67 < z < 1.22)

= P(z < 1.22) - P(z < -0.67 )

Using z table,

= 0.8888 - 0.2514

= 0.6374

= 63.74%

P( 400 < x < 600) = P[(400 - 490)/ 90 ) < (x - ) /  < ( 600 - 490) / 90) ]

= P( -1 < z < 1.22)

= P(z < 1.22 ) - P(z < -1 )

Using z table,

= 0.8888 - 0.1587

= 0.7301

probability = 0.7301