The tensile strength of paper is modeled by a normal distribution with a mean of 35 pounds per square inch and a standard deviation of 2 pounds per square inch. Required:
a. What is the probability that the strength of a sample is less than 38 lb/in²?
b. If the specifications require the tensile strength to exceed 30 lb/in², what proportion of the samples is scrapped?

a) the probability that the strength of a sample is less than 38 lb/in² is approximately 0.9332

b) approximately 0.62% of the sample are scrapped

Step-by-step explanation:

a)

What is the probability that the strength of a sample is less than 38 lb/in²?

lets X rep the paper tensile strength in pounds per square inch.

the probability that the strength of the sample is less than 38 lb/in²

i.e P( X < 38 )

so assuming that X is normally distributed with u = 35 lb/in² and α = 2 lb/in²,

so the probability expression P( X < 38 ) can be standardized as

P( X < 38 ) = P( (X-35)/2  < (38-35)/2)  

= P( X < 1.5 )  

from the cumulative standard normal distribution table,

we find P( X < 1.5 )  =  0.9332

so

P( X < 38 ) = P( X < 1.5 )  

= 0.9332

therefore the probability that the strength of a sample is less than 38 lb/in² is approximately 0.9332

b)

If the specifications require the tensile strength to exceed 30 lb/in², what proportion of the samples is scrapped?

this can be represented as P( X < 30 )

By standardization the probability expression  P( X < 30 ) can be transformed as;

P( X < 30 ) = P( (X-35)/2 < (30-35)/2)  

=  P( z < -2.5 )

from the cumulative standard normal distribution table,

we find P( X < -2.5 )  =  0.0062

so

P( X < 30 ) = P( X < -2.5 )  

= 0.0062

therefore, approximately 0.62% of the sample are scrapped

5x+40

5(x + 8)

7x - 49

7(x - 7)

18x + 9

9(2x + 1)

15x - 10

5(3x - 2)

step-by-step explanation:

for number one all you have to do is factor 5 out of 5x + 40

for number 2 factor 7 out of 7x - 49

for number three factor 9 out of 18x + 9

for number four factor 5 out of 15x - 10

32

step-by-step explanation: