Answer :
Answer:
[tex]P(\bar X<67.2)=P(\frac{X-\mu}{\sigma_{\bar X}}<\frac{23.8-\mu}{\sigma})=P(Z<\frac{23.8-24.3}{0.4526})=P(z<-1.11)[/tex]
And we can find this probability using the normal standard distribution or excel and we got:
[tex]P(z<-1.105)=0.1335[/tex]
Step-by-step explanation:
Let X the random variable that represent the lifetimes of cell phones, and for this case we know the distribution for X is given by:
[tex]X \sim N(24.3,2.6)[/tex]
Where [tex]\mu=24.3[/tex] and [tex]\sigma=2.6[/tex]
We select a sample size of n = 33 emplloyees and we are interested on this probability
[tex]P(\bar X<23.8)[/tex]
And the best way to solve this problem is using the normal standard distribution and the z score given by:
[tex]z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
The standard deviation would be:
[tex]\sigma_{\bar X} = \frac{2.6}{\sqrt{33}}= 0.4526[/tex]
If we apply this formula to our probability we got this:
[tex]P(\bar X<67.2)=P(\frac{X-\mu}{\sigma_{\bar X}}<\frac{23.8-\mu}{\sigma})=P(Z<\frac{23.8-24.3}{0.4526})=P(z<-1.11)[/tex]
And we can find this probability using the normal standard distribution or excel and we got:
[tex]P(z<-1.105)=0.1335[/tex]