A radio manufacturer believes that the length of life of WPPX model radio is normal with µ=12 years and σ=2.5 years. (a) What percent of the radios will function for more than 16 years? (b) Suppose the company decides to replace 1% of the radios. Find the length of the guarantee period. i. e. find X. (c) What percent of the radios that will fail to satisfy the guarantee period of 10 years? , i. e. less than 10 years? (d) If 10% of the radios will function for more than X years, find X. Help, please!!

(a) 5.48%

(b) 17.82 years

(c) 21.19%

(d) 15.21 years

Step-by-step 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