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3 years ago

The amount of toothpaste in a tube is normally distributed with a mean of 6.5 ounces and a standard

Mathematics

1 answer:

Anarel [89]3 years ago

5 0

Answer:

a) 266 tubes , TC_r = $53.2

b) 266 tubes , T.Loss = $13.30

Step-by-step explanation:

Given:

- The sample size of tubes n = 1,000 tubes

- The mean of the sample u = 6.5 oz

- The standard deviation of the sample s.d = 0.8 oz

- Cost of manufacturing a tube C_t = 50 cents

- Cost of refilling a tube C_r = 20 cents

- Profit loss per tube Loss = 5 cents

Find:

a). How many tubes will be found to contain less than 6 ounces? In that case, what will be the total cost of the refill?

b) How many tubes will be found to contain more than 7 ounces? In that case, what will be the amount of profit lost?

Solution:

- First we will compute the probability of tube containing less than 6 oz.

- Declaring X : The amount of toothpaste.

Where, X ~ N ( 6.5 , 0.8 )

- We need to compute P ( X < 6 oz )?

Compute the Z-score value:

P ( X < 6 oz ) = P ( Z < (6 - 6.5) / 0.8 ) = P ( Z < -0.625 )

Use the Z table to find the probability:

P ( X < 6 oz ) = P ( Z < -0.625 ) = 0.266

- The probability that it lies below 6 ounces. The total sample size is n = 1000.

The number of tubes with X < 6 ounces = 1000* P ( X < 6 oz )

= 1000*0.266 = 266 tubes.

- The total cost of refill:

TC_r = C_f*(number of tubes with X < 6)

TC_r = 20*266 = 5320 cents = $53.2

- We need to compute P ( X > 7 oz )?

Compute the Z-score value:

P ( X > 7 oz ) = P ( Z > (7 - 6.5) / 0.8 ) = P ( Z < 0.625 )

Use the Z table to find the probability:

P ( X > 7 oz ) = P ( Z > 0.625 ) = 0.266

- The probability that it lies above 7 ounces. The total sample size is n = 1000.

The number of tubes with X > 7 ounces = 1000* P ( X > 7 oz )

= 1000*0.266 = 266 tubes.

- The total cost of refill:

T.Loss = Loss*(number of tubes with X > 7)

T.Loss = 5*266 = 1330 cents = $13.30

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