Answer:
1) If the manufacturer's assumptions are correct, it would reed to replace 0.621% of its batteries free
2) The company finds that it is replacing 1.07% of its batteries free of charge. It suspects that its assumption standard deviation of the life of its batteries is incorrect. A standard deviation of 6.084 results in a 1.07% replacement rate
3) Using the revised standard deviation for battery life, the percentage of the manufacturer's batteries that don't qualify for free replacement but do qualify for the prorated credit = 92%
Step-by-step explanation:
This is a normal distribution problem with
Mean = μ = 45 months
Standard deviation = σ = 5.6 months.
Batteries that fail within the first 31 months get a free replacement and batteries that fail after 31 months but within 54 months get a prorated credit toward the purchase of a new battery.
1) Percentage of batteries that'll qualify for a replacement are the batteries that fail within 31 months. That is, P(X ≤ 31)
To obtain this, we first normalize or standardize 31 months.
The standardized score for any value is the value minus the mean then divided by the standard deviation.
z = (x - μ)/σ = (31 - 45)/5.6 = - 2.50
The required probability
P(X ≤ 31) = P(z ≤ -2.50)
We'll use data from the normal probability table for these probabilities
P(X ≤ 31) = P(z ≤ -2.50) = 0.00621 = 0.621%
2) The company finds that the percentage of free batteries replaced is actually 1.07%
P(X ≤ 31) = 1.07% = 0.0107
Let the z value of 31 months under this new normal distribution be z'
P(X ≤ 31) = P(z ≤ z') = 0.0107
From the normal distribution table,
z' = -2.301
z' = (x - μ)/σ
-2.301 = (31 - 45)/σ
σ = (-14) ÷ (-2.301) = 6.084 months
3) Using the revised standard deviation for battery life, what percentage of the manufacturer's batteries don't free replacement but do qualify for the prorated credit?
That is, P(31 < X ≤ 54)
We first normalize 31 and 54 using the revised standard deviation
For 31
z = (x - μ)/σ = (31 - 45)/6.084 = - 2.30
For 54
z = (x - μ)/σ = (54 - 45)/6.084 = 1.48
The required probability = P(31 < X ≤ 54)
= P(-2.30 < z ≤ 1.48)
Using the normal distribution tables
P(31 < X ≤ 54) = P(-2.30 < z ≤ 1.48)
= P(z ≤ 1.48) - P(z < -2.30)
= 0.93056 - 0.0107
= 0.91986 = 91.986% = 92% to nearest whole number.
Hope this Helps!!!!
