Question

Suppose that the weight of pears from California is normally distributed with a mean of 9 ounces and a standard deviation of 1.5 ounces.

a. What is the probability of selecting a pear that weighs more than 12 ounces?

b. What is the probability of selecting a pear that weighs less than 6 ounces?

c. What is the probability of selecting a pear that weigh exactly 10.5 ounces?

d. What is the probability of selecting a pear that weighs between 7.5 and 10.5 ounces?

Answer 1

Solution :

Given that,

mean = = 9

standard deviation = = 1.5

a ) P (x > 12 )

= 1 - P (x < 12 )

= 1 - P ( x -  / ) < ( 12 - 9 / 1.5)

= 1 - P ( z < 3 / 1.5 )

= 1 - P ( z < 2 )

Using z table

= 1 - 0.9772

= 0.0228

Probability = 0.0228

b ) P( x < 6 )

P ( x - / ) < ( 6 - 9 / 1.5)

P ( z < -3 / 1.5 )

P ( z < - 2 )

= 0.0228

Probability = 0.0228

c ) P( x = 10.5 ) = 0

d ) P (7.5 < x < 10.5 )

P ( 7.5 - 9 / 1.5) < ( x -  / ) < ( 10.5 - 9 / 1.5 )

P ( - 1.5 / 1.5 < z < 1.5 / 1.5 )

P (-1 < z < 1 )

P ( z < 1 ) - P ( z < -1)

Using z table

= 0.8413 - 0.1587

= 0.6826

Probability = 0.6826