Respuesta :
Answer:
[tex] z= \frac{9832- 10000}{\frac{714.2857}{\sqrt{49}}}= -1.646[/tex]
[tex] z= \frac{10200- 10000}{\frac{714.2857}{\sqrt{49}}}= 1.96[/tex]
And we can use the normal standard distribution table or excel and we can find the probability with this difference:
[tex] P(-1.646 <z< 1.96) =P(z<1.96) -P(z<-1.646) =0.975-0.0499= 0.9251[/tex]
Then the probability that the sample mean breaking strength for a random sample of 49 rivets is between 9,832 and 10,200 is 0.9251
Step-by-step explanation:
For this case we have the following info given:
[tex] \mu = 10000[/tex] represent the mean
[tex] \sigma = 714.2857[/tex] represent the deviation
[tex] n = 49[/tex] represent the sample size selected
For this case since the sample size is large enough n>30 we have enough evidence to use the central llmit theorem and the distribution for the sample mena would be given by:
[tex] \bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}}) [/tex]
And we want to find the following probability:
[tex] P(9832 < \bar X< 10200)[/tex]
And we can use the z score formula given by:
[tex] z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And if we use the z score formula for the limits given we got:
[tex] z= \frac{9832- 10000}{\frac{714.2857}{\sqrt{49}}}= -1.646[/tex]
[tex] z= \frac{10200- 10000}{\frac{714.2857}{\sqrt{49}}}= 1.96[/tex]
And we can use the normal standard distribution table or excel and we can find the probability with this difference:
[tex] P(-1.646 <z< 1.96) =P(z<1.96) -P(z<-1.646) =0.975-0.0499= 0.9251[/tex]
Then the probability that the sample mean breaking strength for a random sample of 49 rivets is between 9,832 and 10,200 is 0.9251
