Question

Suppose the life of a particular brand of calculator battery is approximately normally distributed with a mean of 75 hours and a standard deviation of 9 hours. Complete parts a through c.

a. What is the probability that a single battery randomly selected from the population will have a life between 70 and 80 hours? P(70 < or = x overbar < or = 80) = 0.4246 (Round to four decimal places as needed.)

b. What is the probability that 4 randomly sampled batteries from the population will have a sample mean life of between 70 and 80 hours? P(70 < or = x overbar < or = 80) = (Round to four decimal places as needed.

Answer #1

Solution:

Part a

We are given

µ = 75

σ =9

We have to find P(70 < X < 80)

P(70 < X < 80) = P(X<80) – P(X<70)

Find P(X<80)

Z = (X - µ)/σ

Z = (80 – 75)/9

Z = 0.555555556

P(Z< 0.555555556) = P(X<80) = 0.710742639

(by using excel/z-table)

Now find P(X<70)

Z = (X - µ)/σ

Z = (70 – 75)/9

Z = -0.555555556

P(Z< -0.555555556) = P(X<70) = 0.289257361

(by using excel/z-table)

P(70 < X < 80) = P(X<80) – P(X<70)

P(70 < X < 80) = 0.710742639 - 0.289257361

P(70 < X < 80) = 0.421485278

Required probability = 0.4215

Part b

We are given

µ = 75

σ =9

n = 4

We have to find P(70 < Xbar < 80)

P(70 < Xbar < 80) = P(Xbar<80) – P(Xbar<70)

Find P(Xbar<80)

Z = (X - µ)/[σ/sqrt(n)]

Z = (80 – 75)/[9/sqrt(4)]

Z = 1.111111

P(Z< 1.111111) = P(Xbar<80) = 0.86674

(by using excel/z-table)

Now find P(Xbar<70)

Z = (X - µ)/[σ/sqrt(n)]

Z = (70 – 75)/[9/sqrt(4)]

Z = -1.111111

P(Z< -1.111111) = P(Xbar<70) = 0.13326

(by using excel/z-table)

P(70 < Xbar < 80) = P(Xbar<80) – P(Xbar<70)

P(70 < Xbar < 80) = 0.86674 - 0.13326

P(70 < Xbar < 80) = 0.73348

**Required probability =**
**0.7335**

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