Answer:
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.
