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The yield stress of a random sample of 25 pieces of steel was measured, yielding a mean of 52,800 psi. and an estimated standard deviation of s = 4,600 psi. a. What is the probability that the population mean is less than 50,000 psi? b. What is the estimated fraction of pieces with yield strength less than 50,000 psi? c. Is this sampling procedure sampling-by-attributes or sampling-by-variable?

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fortuneonyemuwa

Answer:

Step-by-step explanation:

Given than n = 25 , mean = 52800 sd = 4600

1) P(X<50000) , so please keep the z tables ready

we must first convert this into a z score so that we can look for probability values in the z table

using the formula

Z = (X-Mean)/SD

(50000- 52800)/4600 = -0.608

checkng the value in the z table

P ( Z>−0.608 )=P ( Z<0.608 )=0.7291

b)

again using the same formula and converting to z score

we need to calculate P(X<50000)

Z = (X-Mean)/SD

(50000- 52800)/4600 = -0.608

P ( Z<−0.608 )=1−P ( Z<0.608 )=1−0.7291=0.2709

proportion is 27% approx

c)

When your data points are measurements on a numerical scale you have variables data , here yield stress is numeric in nature , hence its a sampling by variable plan

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