Question

14.

Answer 1

Answer:

0.8945

Step-by-step explanation:

The average life of smokers, $\mu$ is 60 months

Standard deviation is 8 months

Random sample, n is 30 smoke detectors

$P(58 < \bar x < 63)$

= $P((58 - \mu)/(\sigma/\sqrt n) < (\bar x - \mu)/(\sigma/\sqrt n) < (63 - \mu)/(\sigma/\sqrt n))$

= $P((58 - 60)/(8/\sqrt 30) < Z < (63 - 60)/(8/\sqrt 30))$

= P(-1.37 < Z < 2.05)

= P(Z < 2.05) - P(Z < -1.37)  Then from the standard tables, the Z values are 0.9798 for Z=2.05 and 0.0853 for Z=-1.37

= 0.9798 - 0.0853

= 0.8945

Therefore, the probability that a random sample of 30 smoke detectors will have a lifetime between 58 months and 63 months is 0.8945