A certain type of thread is manufactured with a mean tensile strength of 76.7 ki

Question

A certain type of thread is manufactured with a mean tensile strength of 76.7 kilograms and a standard deviation of 5.9 kilograms. Assume a sample of 100 items is obtained.

(a) Do we know the distribution of the tensile strength of any produced item?

(b) What is the mean and standard deviation of the sample average?

(c) What can you say about the distribution of the sample average?

(d) Given your answer in part (c), find the probability that a sample of 100 items will have an average

i. greater than 78.

ii. less tan 75.

iii. between 75.8 and 76.8.

Explanation / Answer

We dont know the relationship between tensile strength and thread , hence this question cant be answered here in statistics form. Check this in Menchanical engg forum

(b)
The mean of the sampling distribution will also be 76.7 kg, but the standard deviation of the distribution of samples of 100 will be S/sqrt(N) , where n is the sample size in this case 100

(5.9 kg)/sqrt(100) = 0.59 kg.

(c) If the population is normally distributed, the distribution of the means of samples of 100 will also be normally distributed.

d) Now

P(X>78) , we can convert this to z score using (X-Mean)/SD as our formula , please keep the z tables handy for this

Z= (78 - 76.7)/0.59 = 2.203 , P(Z>2.203)

so we get

P ( Z>2.203 )=1P ( Z<2.203 )=10.9861=0.0139

ii) P(X<75)

again using the same formula

(75-76.7)/0.59 = -2.881

P ( Z<2.881 )=1P ( Z<2.881 )=10.998=0.002

iii) P(75.8<x<76.8)

so we get

(75.8-76.7)/0.59 <z< (76.8-76.7)/0.59

75.8 corresponds to z = -1.525
76.8 corresponds to z = 0.1695

so the value can be calculated as

To find the probability of P (1.525<Z<0.1695), we use the following formula:

P (1.525<Z<0.1695 )=P ( Z<0.1695 )P (Z<1.525 )

P ( Z<1.525 ) can be found by using the following fomula.

P ( Z<a)=1P ( Z<a )

After substituting a=1.525 we have:

P ( Z<1.525)=1P ( Z<1.525 )

We see that P ( Z<1.525 )=0.937 so,

P ( Z<1.525)=1P ( Z<1.525 )=10.937=0.063

At the end we have:

P (1.525<Z<0.1695 )=0.5045

Hope this helps !!

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