Mercury contamination of swordfish. Consumer Reports 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. Use the normal approximation to the binomial to calculate the probability that fewer than 2 of the 20 swordfish pieces have mercury levels exceeding the FDA limit. (The final answer is 0.0015) b. Use the normal approximation to the binomial to calculate the probability that more than half of the 20 swordfish pieces have mercury levels exceeding the FDA limit. (The final answer is 0.1271) c. Use the binomial tables to calculate the exact probabilities in parts a and b. Does the normal distribution provide a good approximation to the binomial distribution? (0.0005 and 0.1275)

Using the normal approximation to the binomial distribution

a) The probability that fewer than 2 of the 20 swordfish pieces have mercury levels exceeding the FDA limit = P(x < 2) = 0.0015

b) The probability that more than half of the 20 swordfish pieces have mercury levels exceeding the FDA limit = P(x > 10) = 0.1271

Using Binomial distribution formula for these same probabilities.

c) The probability that fewer than 2 of the 20 swordfish pieces have mercury levels exceeding the FDA limit = P(x < 2) = 0.0005

The probability that more than half of the 20 swordfish pieces have mercury levels exceeding the FDA limit = P(x > 10) = 0.1275

With the continuity correction factor added, the normal approximation to the binomial distribution problem is a good estimate of the binomial distribution, especially for variables close to the mean.

Step-by-step explanation:

To use the normal approximation to the binomial distribution problem, we need to compute the mean and the standard deviation.

With n = sample size = 20

p = proportion of the swordfish pieces available for sale have a level of mercury above the Food and Drug Administration (FDA) limit = 40% = 0.40

Mean = μ = np = 20 × 0.40 = 8

Standard deviation = σ = √[np(1-p)] = √(20×0.40×0.60) = 2.191

If X represents the number of swordfish pieces that have mercury levels exceeding the FDA limit

a) Probability that fewer than 2 of the 20 swordfish pieces have mercury levels exceeding the FDA limit.

Introducing the continuity correction factor,

P(X < 2) becomes P(x < 2-0.5) = P(x < 1.5)

We first normalize or standardize 1.5.

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ = (1.5 - 8)/2.191 = - 2.97

To determine the required probability

P(x < 1.5) = P(z < -2 97)

We'll use data from the normal distribution table for these probabilities

P(X < 2) = P(x < 1.5) = P(z < -2 97) = 0.00149 = 0.0015

b) Probability that more than half of the 20 swordfish pieces have mercury levels exceeding the FDA limit

Adjusting with the continuity correction factor P(X > 10) = P(x > 10+0.5) = P(x > 10.5)

We first normalize or standardize 10.5

z = (x - μ)/σ = (10.5 - 8)/2.191 = 1.14

To determine the required probability

P(x > 10.5) = P(z > 1.14)

We'll use data from the normal distribution table for these probabilities

P(X > 10) = P(X > 10.5) = P(z > 1.14)

= 1 - P(z ≤ 1.14)

= 1 - 0.87286 = 0.12714 = 0.1271

c) Binomial distribution function is represented by

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = number of swordfishes to be examined = 20

x = Number of successes required = fewer than 2 and more than half, < 2 and > 10

p = probability of success = probability of a swordfish having mercury levels above the FDA limits = 0.40

q = probability of failure = probability of a swordfish NOT having mercury levels above the FDA limits = 1 - p = 1 - 0.40 = 0.60

P(X < 2) = P(X = 0) + P(X = 1) = 0.00052404938 = 0.0005

P(X > 10) = P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20) = 0.12752124615 = 0.1275

With the continuity correction factor added, the normal approximation to the binomial distribution problem is a good estimate of the binomial distribution, especially for variables close to the mean.

Hope this Helps!!!

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step-by-step explanation:

answer: about 22.67

step-by-step explanation: