A company that manufactures mufflers for cars offers a lifetime warranty on its products, provided that ownership of the car does not change. Only 20% of its mufflers are replaced under this warranty.

a. Define the random variable X as the number of mufflers replaced in a sample of 400 purchases. Give the distribution of X.
b. Referring to part 2a), what is the approximate probability that between 75 and 100 mufflers (inclusive) arc replaced under warranty?
c. Among 400 randomly selected purchases, what is the approximate probability that at most 70 are ultimately replaced under warranty?

a) Solution is presented in the attached image.

b) 0.728

c) 0.106

Step-by-step explanation:

First of, we check the type of distribution this is,

Mean = xbar = np = 400 × 0.2 = 80

Standard deviation = √[np(1-p)] = √(80×0.8) = 8

Condition for the distribution to be a normal distribution

np ≥ 10

80 ≥ 10

And

np(1-p) ≥ 10

80(0.8) = 320 ≥ 10

Hence, this is a normal distribution

a) For normal distribution, the distribution of X is given by the equation attached to this solution

F(x) = sum of probabilities up to the point of x.

b) The approximate probability that between 75 and 100 (inclusive) mufflers are replaced under warranty?

Normal distribution can be solved using formula or using the tables.

To use the formula, we just plug the values of mean (μ), standard deviation (σ) and the variable whose probability is required (x) in the equation in (a) and obtain the solutions.

But, I'll use the tables, so that this would be clearer.

We first convert the values 75 and 100 to standardized scores.

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

z = (x - μ)/σ

For 75

z = (75 - 80)/8 = - 0.625

For 100

z = (100 - 80)/8 = 2.5

To find the approximate probability that between 75 and 100 (inclusive) mufflers are replaced under warranty

P(75 ≤ x ≤ 100) = P(-0.625 ≤ z ≤ 2.5)

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

P(75 ≤ x ≤ 100) = P(-0.625 ≤ z ≤ 2.5) = P(z ≤ 2.5) - P(z ≤ -0.625) = 0.994 - 0.266 = 0.728

b) Among 400 randomly selected purchases, what is the probability that at most 70 mufflers are ultimately replaced under warranty?

We convert the value 70 to standardized scores.

z = (x - μ)/σ = (70 - 80)/8 = - 1.25

To find the approximate probability that at most 70 mufflers are replaced under warranty

P(x ≤ 70) = P(z ≤ -1.25)

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

P(x ≤ 70) = P(z ≤ -1.25) = 0.106