Question

It is estimated that 80% of a grapefruit crop is good; the other 20% have rotten centers that cannot be detected unless the
grapefruit are cut open. The grapefruit are sold in sacks of 12. Let r be the number of good grapefruit in a sack.
(a) Make a histogram of the probability distribution of r.

(b) What is the probability of getting no more than one bad grapefruit in a sack? (Round your answer to three decimal
places.)
What is the probability of getting at least one good grapefruit? (Round your answer to three decimal places.)
(c) What is the expected number of good grapefruit in a sack? (Round your answer to one decimal place.)
grapefruit
(d) What is the standard deviation of the r probability distribution? (Round your answer to two decimal places.)
grapefruit

Answer 1

Part a)

X	P ( X )
0	0
1	0.0000
2	0.0000
3	0.0001
4	0.0005
5	0.0033
6	0.0155
7	0.0532
8	0.1329
9	0.2362
10	0.2835
11	0.2062
12	0.0687

Histogram

Part b)

X ~ B ( n = 12 , P = 0.2 )

P ( X <= 1 ) = 0.275

X ~ B ( n = 12 , P = 0.8 )

P ( X >= 1 ) = 1 - P ( X = 0 ) = 1.00 i.e 100%

Part c)

Mean = n * P = ( 12 * 0.8 ) = 9.6

Part d)

Variance = n * P * Q = ( 12 * 0.8 * 0.2 ) = 1.92
Standard deviation = (variance) = (1.92) = 1.39