Answer:
Step-by-step explanation:
given that certain tubes manufactured by a company have a mean lifetime of 800 hours and a standard deviation of 60 hours.
Sample size n =16
Std error of sample mean = [tex]\frac{\sigma}{\sqrt{n} } \\= 15[/tex]
x bar follows N(800, 15)
the probability that a random sample of 16 tubes taken from the group will have a mean lifetime
(a) between 790 and 810 hours,
=[tex]P(790<x<810)\\= P(|z|<0.667)\\= 2*0.248\\= 0.496[/tex]
(b) less than 785 hours
[tex]=P(X<785)\\=P(Z<-1)\\= 0.1584[/tex]
, (c) more than 820 hours,
[tex]=P(X>820)\\=p(Z>1.333)\\= 0.0913[/tex]
(d) between 770 and 830 hours
=[tex]P(|Z|<2)\\= 0.4772*2\\=0.9544[/tex]