Question

Consumer Reports (Feb. 1992) found widespread contamination of seafood in New York and Chicago supermarkets. For example, 40 % of the swordfish pieces available for sale have a level of mercury above the Food and Drug Administration (FDA) limit. Consider a random sample of 20 swordfish pieces from New York and Chicago supermarkets.

(a) Calculate the probability that fewer than 2 of the 20 swordfish pieces have mercury levels exceeding the FDA limit. [3]

(b) Calculate the probability that more than half of the 20 swordfish pieces have mercury levels exceeding the FDA limit.

Answer 1

Answer:

Given,

X ~ Bin(n = 20 , p = 0.40)

mean = np = 20*0.40 = 8

Standard deviation = sqrt(npq) = sqrt(20*0.40*0.60) =2.191

a)

Consider,

P(X < 1.5) [since using continuity correction]

= P((x - u)/s < (1.5 - 8)/2.191)

= P(z < -2.97)

= 0.001489 [since from z table]

= 0.0015

Now P(X < 2) = P((x-u)/s < (2 - 8)/2.191)

= P(z < - 2.74)

= 0.0030719 [since from z table]

= 0.0031

b)

Given,

x = 10.5

mean = 8

standard deviation = 2.191

P(X > 10.5) = P((x-u)/s > (10.5 - 8)/2.191)

= P(z > 1.14)

= 0.1271432 [since from z table]

= 0.1271

P(X > 10) = P((x-u)/s > (10 - 8)/2.191)

= P(z > 0.91)

= 0.1814113 [since from z table]

= 0.1814