Question

An expensive watch is powered by a​ 3-volt lithium battery expected to last five years. Suppose the life of the battery has a standard deviation of 0.6 year and is normally distributed.

a. Determine the probability that the​ watch's battery will last longer than 5.7 years.

b. Calculate the probability that the​ watch's battery will last more than 4.7 years.

c. Compute the​ length-of-life value for which ​5% of the​ watch's batteries last longer.

Answer 1

$\mu$=5

$\sigma$=0.6

x~N($\mu$=5,$\sigma$=0.6)

Ansa:

P ( X > 5.7 ) = 1 - P ( X < 5.7 )
Standardizing the value
Z = ( X - µ ) / σ

P ( ( X - µ ) / σ ) > ( 5.7- 5 ) / 0.6 )
P ( Z > 1.1667 )
P ( X > 5.7 ) = 1 - P ( Z < 1.1667 )
P ( X > 5.7 ) = 1 - 0.8784
P ( X > 5.7 ) = 0.1217

Ansb)

P ( X > 4.7 ) = 1 - P ( X < 4.7 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 4.7 - 5 ) / 0.6
Z = -0.5
P ( ( X - µ ) / σ ) > ( 4.7 - 5 ) / 0.6 )
P ( Z > -0.5 )
P ( X > 4.7 ) = 1 - P ( Z < -0.5)
P ( X > 4.7) = 1 - 0.3085
P ( X > 4.7 ) = 0.6915

Ansc)

X ~ N ( µ = 5 , σ = 0.6 )
P ( X > x ) = 1 - P ( X < x ) = 1 - 0.05 = 0.95
To find the value of x
Looking for the probability 0.95 in standard normal table to calculate Z score = 1.6449
Z = ( X - µ ) / σ
1.6449 = ( X - 5 ) / 0.6
X = 5.9869