Answer:
(a) 5.48%
(b) 17.82 years
(c) 21.19%
(d) 15.21 years
Explanation:
z = (x-μ)/σ,
where
x is the raw score
μ is the population mean
σ is the population standard deviation.
µ=12 years and σ=2.5 years.
(a) What percent of the radios will function for more than 16 years?
For x = 16
z = 16 - 12/2.5
z = 1.6
Probability value from Z-Table:
P(x<16) = 0.9452
P(x>16) = 1 - P(x<16)
1 - 0.9452
= 0.054799
Converting to percentage = 0.054799 × 100
= 5.4799%
Approximately = 5.48%
(b) Suppose the company decides to replace 1% of the radios. Find the length of the guarantee period. i.e. find X.
find the z score of the 99th percentile = 2.326
Hence:z = (x-μ)/σ
2.326 = x - 12/2.5
Cross Multiply
2.326 × 2.5 = x - 12
5.815 = x - 12
x = 12 + 5.815
x = 17.815
Approximately = 17.82 years
(c) What percent of the radios that will fail to satisfy the guarantee period of 10 years? , i.e. less than 10 years?
When x < 10
Hence,
z = 10 - 12/2.5
z = -0.8
Probability value from Z-Table:
P(x<10) = 0.21186
Converting to percentage
0.21186 × 100
= 21.186%
Approximately = 21.19%
(d) If 10% of the radios will function for more than X years, find X.
The z score would be : 100 - 10%
= 90th percentile z score
We find the z score of the 90th percentile = 1.282
Hence:z = (x-μ)/σ
1.282 = x - 12/2.5
Cross Multiply
1.282 × 2.5 = x - 12
3.205 = x - 12
x = 12 + 3.205
x = 15.205
Approximately = 15.21 years