Question

. Suppose that banana prices at supermarkets in a particular large metropolitan area follow a normaldistribution with mean µ= 58.6 cents per pound and standard deviation σ= 4.3 cents per pound.

For parts (a) through (c), what is the probability that the price of bananas at a random supermarket will be:

(a)  below 55 cents per pound?

(b)  above 65 cents per pound?  

(c)  Between 57 and 62 cents per pound?

(d) Below 57 cents or above 62 cents per pound?

(d)  What is the 25^th percentile of banana prices in this particular area?

Answer 1

Solution :

Given that ,

mean = = 58.6

standard deviation = = 4.3

a) P(x < 55) = P[(x - ) / < (55 -58.6) /4.3 ]

= P(z < -0.84 )

= 0.2005

probability =0.2005

b)

P(x > 65) = 1 - p( x<65 )

=1- p [(x - ) / < (65 -58.6) /4.3 ]

=1- P(z < 1.49)

= 1 - 0.9319 = 0.0681

probability = 0.0681

c)

P( 57< x < 62 ) = P[(57 -58.6)/4.3 ) < (x - ) /  < (62 - 58.6) /4.3 ) ]

= P(-0.37 < z < 0.79 )

= P(z < 0.79 ) - P(z < -0.37 )

Using standard normal table

= 0.7852 - 0.3557 = 0.4295

Probability = 0.4295

d)

P(x < 57) = P[(x - ) / < (57 -58.6) /4.3 ]

= P(z < -0.37 )

= 0.3557

P(x > 62) = 1 - p( x<62 )

=1- p [(x - ) / < (62 -58.6) /4.3 ]

=1- P(z < 0.79)

= 1 - 0.7852= 0.2148

probability = 0.3557 +0.2148 = 0.5705

e)

P(Z < z) = 0.25

z =-0.674

Using z-score formula,

x = z * +

x = -0.674 * 4.3+58.6

x = 55.7