The tensile strength of a certain metal component is normally distributed with a mean of 10,000 kilograms per square centimeter and a standard deviation of 100 kilograms per square centimeter. Measurements are recorded to the nearest 50 kilograms per square centimeter.

Required:
a. What proportion of these components exceed 10,150 kilograms per square centimeter in tensile strength?
b. If specifications require that all components have tensile strength between 9800 and 10,200 kilograms per square centimeter inclusive, what proportion of pieces would we expect to scrap?

Question

contestada

Respuesta :

jmonterrozar jmonterrozar · Answer 1 · 2020-07-06T03:53:29+02:00

Answer:

a. 0.0668

b. 0.9545

Step-by-step explanation:

We have the following information:

mean (m) = 10000

standard deviation (sd) = 100

(a)

We must calculate the proportion of the components exceed 10150 kilograms per square centimeter in tensile strength as follows:

P (x> 10150) = P [(x - m) / sd> (10150 - 1000 /) 100]

P (x> 10150) = P (z> 1.5)

P (x> 10150) = 1 - P (z <1.5)

P (x> 10150) = 1 - 0.9332 (attached table)

P (x> 10150) = 0.0668

Therefore the proportion of the components exceed 10150 kilograms per square centimeter in tensile strength is 0.0668

(b)

We must calculate the proportion of all components has tensile strength between 9800 and 10200, as follows:

P (9800 <x <10200) = P [(9800 - 1000 /) 100 <(x - m) / sd <(10200 - 1000 /) 100]

P (9800 <x <10200) = P (-2 <z <2)

P (9800 <x <10200) = P (z <2) - P (z <-2)

P (9800 <x <10200) = 0.9773 - 0.0228 (attached table)

P (9800 <x <10200) = 0.9545

the proportion of pieces that would expect to scrap is 0.9545