Which point on the x axis lies on the line that passes through point c and is parallel to line ab

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!!!