When the parameters of a population are known, the likelihood of obtaining a certain mean value, ¯ x , from a small sample can be predicted by the z‐variable with z = ( ¯ x − x ' ) / σ / √ N . Consider a stamping process in which the applied load has a known true mean of 400 N with standard deviation of 25 N. If 25 measurements of applied load are taken at random, what is the probability that this sampling will have a mean value between 390 and 410?

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JeanaShupp JeanaShupp · Answer 1 · 2019-09-26T05:54:54+02:00

Answer: 0.9545

Step-by-step explanation:

Given : Population mean : [tex]x'=400\ N[/tex]

Standard deviation : [tex]\sigma=25\ N[/tex]

Sample size : n=25

Test statistic : [tex]z=\dfrac{\overline{x}-x'}{\dfrac{\sigma}{\sqrt{n}}}[/tex]

For [tex]\overline{x}= 390[/tex], we have

[tex]z=\dfrac{390-400}{\dfrac{25}{\sqrt{25}}}=-2[/tex]

For [tex]\overline{x}= 410[/tex], we have

[tex]z=\dfrac{410-400}{\dfrac{25}{\sqrt{25}}}=2[/tex]

Now, by using the standard normal distribution table for z , we have

[tex]\text{P-value=}P(-2<z<-2)=1-2(P\geq2)\\\\=1-2(1-P(z<2))\\\\=1-2(1-0.9772498)=0.9544996\approx0.9545[/tex]

Hence, probability that this sampling will have a mean value between 390 and 410 = 0.9545