A company that manufactures mufflers for cars offers a lifetime warranty on its products, provided that ownership of the car does not change. Suppose that only 20% of its mufflers are replaced under this warranty. a. In a random sample of 400 purchases, what is the approximate probability that between 75 and 100 (inclusive) mufflers are replaced under warranty? b. Among 400 randomly selected purchases, what is the probability that at most 70 mufflers are ultimately replaced under warranty? c. If you were told that fewer than 50 among 400 randomly selected purchases were ever replaced under warranty, would you question the 20% figure? Explain.

Question

Mathematics

Tyrese Mays

13 May, 07:22

A company that manufactures mufflers for cars offers a lifetime warranty on its products, provided that ownership of the car does not change. Suppose that only 20% of its mufflers are replaced under this warranty. a. In a random sample of 400 purchases, what is the approximate probability that between 75 and 100 (inclusive) mufflers are replaced under warranty? b. Among 400 randomly selected purchases, what is the probability that at most 70 mufflers are ultimately replaced under warranty? c. If you were told that fewer than 50 among 400 randomly selected purchases were ever replaced under warranty, would you question the 20% figure? Explain.

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Answers (1)

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Harley Meyers · Answer 1

a) 0.728

b) 0.106

c) Yes,

The mean given needs to be seriously questioned because, for a normal distribution, all of the possible values (95% of the values) should lie within the range of z = - 3 and z = 3. The value for z = - 3.75 is an outlier and shouldn't happen for a normal distribution.

Step-by-step explanation:

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, we use the z-tables for this

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

We 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 - xbar) / 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 7 to standardized scores.

z = (x - xbar) / 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

c) If you were told that fewer than 50 among 400 randomly selected purchases were ever replaced under warranty, would you question the 20% figure? Explain.

We first convert 50 to standard scores

z = (x - xbar) / σ = (50 - 80) / 8 = - 3.75

The mean given needs to be seriously questioned then, because for a normal distribution, all of the possible values (95%) should lie within the range of z = - 3 and z = 3. The value for z = - 3.75 is an outlier for normal distribution.