What is the length of TW? Please help me with this.

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 pe

kicyunya [14]

Answer:

(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

4 0

3 years ago

10-kg bag of cherries for $8<br><br> ___ per kg

Sophie [7]

$8.00/10 kg equals $0.80/kg

7 0

3 years ago

What fraction of two pounds is fifty pence?

EastWind [94]

1/4

There are 100 pence in a pound, so 2 pounds is 200 pence. 50 is 1/4 of 200.

5 0

3 years ago

What percent of 75 is 25

nikdorinn [45]

P / 100 = 25 / 75;

p = 2500 / 75 ;

p = 33.33;

p% = 33.33%

4 0

3 years ago

A computer system uses passwords that are six characters, and each character is one of the 26 letters (a–z) or 10 integers (0–9)

Blababa [14]

First of all, since we have 36 characters available per spot (26 letters and 10 digits), and we have 6 spots, we have a total of

$36^6$

possible passwords.

Event A happens if the password starts with either a, e, i, o or u. If we fix the first character, we're left with 36 characters available for each of the remaining 5 spots, leading to a total of

$5\cdot 36^5$

possible passwords.

So, the probability of event A, computed as the ratio between "good" cases and all possible cases, is

$\dfrac{5\cdot 36^5}{36^6}=\dfrac{5}{36}$

Event B works exactly the same, since we're fixing the last spot, leaving 36 characters available for each of the first 5 spots. So, we have

$P(A)=P(B)=\dfrac{5}{36}$

As for the intersection, we want the first character to be a vowel, and the last character to be an even digits. There are 25 passwords satisfying this request:

$axxxx0,\ axxxx2,\ axxxx4,\ axxxx6,\ axxxx8$