Question

The shear strength of each of ten test spot welds is determined, yielding the following data (psi).

Answer 1

Answer:

a) Mean = 382.3 psi

Standard deviation = 20.8 psi

b) 392.7 psi

c) P(X<400)=0.8026

Step-by-step explanation:

a) The population mean can be estimated as equal to the mean of the sample, and the population standard deviation can be estimated from the sample standard deviation:

$M=\frac{1}{n}\sum x_i=\frac{1}{10}(389+405+409+367+358+415+376+375+367+362)\\\\M=\frac{1}{10}(3823)=382.3\\\\\mu\approx M=382.3$

$s=\sqrt{\frac{1}{n-1}\sum (x_i-M)^2}=\sqrt{\frac{1}{10-1}(389-382.3)^2+...+(362-383.2)^2}\\\\\\s=\sqrt{\frac{1}{9}(3906.1)}=\sqrt{434}=20.8\\\\\\\sigma\approx s=20.8$

b) We start by searching for the z-value for the 95th percentile. This value is

z=1.645:

$P(z$

Then, the strength value below which 95% of all welds will have their strengths is:

$X=\mu+z\cdot \sigma/\sqrt{n}=382.3+1.645*20.8/\sqrt{10}\\\\X=382.3+32.9/3.2=382.3+10.4\\\\X=392.7$

c) We calculate the probability of X being equal or less than 400 as:

$z=(X-\mu)/\sigma=(400-382.3)/20.8=17.7/20.8=0.851\\\\\\P(X\leq400)=P(z$