Question

Melons from a certain large distributor have diameters that follow an approximately normal distribution with a mean of 133 millimeters (mm) and a standard deviation of 5 mm.

a. If a melon from this distributor is randomly selected, calculate the probability that the melon will have a diameter that is greater than 137 mm.

b. Calculate the diameter of a melon such that 15 percent of this distributor’s melons have a larger diameter.

c. Suppose a customer randomly selects melons from this distributor’s inventory until he obtains a melon with a diameter that is greater than 137 mm. Calculate the probability that the first such melon is the fourth melon that the customer selects.

d. Suppose five melons are randomly selected from this distributor’s inventory and that the diameter of each selected melon is recorded.

1. Calculate the mean and the standard deviation of the sampling distribution of the mean diameter for random samples of five melons.

2. Calculate the probability that the mean diameter for a random sample of five melons is greater than 137 mm.

Answer 1

Solution :

Given that ,

mean = = 133

standard deviation = = 5

a) P(x > 137 ) = 1 - p( x< 137)

=1- p [(x - ) / < (137 -133) / 5 ]

=1- P(z < 0.80)

= 1 - 0.7881 = 0.2119

probability = 0.2119

b)

P(Z > z ) = 0.15

1- P(z < z) =0.15

P(z < z) = 1-0.15 = 0.85

z =1.036

Using z-score formula,

x = z * +

x = 1.036 * 5+ 133

x = 138.18

c)

n = 4

= = 133  

= / n = 5/ 4 = 2.5

P( > 137 ) = 1 - P( < 137)

= 1 - P[( - ) / < (137 -133) /2.5 ]

= 1 - P(z < 1.6)

= 1 - 0.9452 = 0.0548

Probability = 0.0548

d)

n = 5

Mean = = = 133  

Standrd deviation = = / n = 5/ 5 = 2.2361

P( > 137 ) = 1 - P( < 137)

= 1 - P[( - ) / < (137 -133) /2.2361 ]

= 1 - P(z < 1.79)

= 1 - 0.9633 = 0.0367

Probability = 0.0367