Final answers are a- 51.4% b- 13.57% c- 34.13% Statement: A battery manufacturer

Question

Final answers are

a- 51.4%

b- 13.57%

c- 34.13%

Statement: A battery manufacturer guarantees that his batteries will last, on aver- age, 3 years, with a standard deviation of 1 year. His claims are based upon a very large population of his 'good' batteries. A consumer watch group decides to test his guarantee. Their small sample of his 'good' batteries indicates bat- tery lifetimes (in years) of 1.9, 2.4, 3.0, 3.5, and 4.2. Determine (a) the percent confidence that the difference between the watch group's sample variance and manufacturer's true variance is due solely to random effects. Next, based upon the manufacturer's battery population average life time and standard deviation, (b) determine the probabilities that a battery lifetime will be less than 1.9 years and (c) between 3 and 4 years

Explanation / Answer

Part a

We are given a sample as

1.9, 2.4, 3.0, 3.5, 4.2

From this sample, we have

Sample mean = Xbar = 3.0

Sample variance = S2 = 0.815

Sample standard deviation = S = sqrt(0.815) = 0.902774

Sample size = n = 5

Degrees of freedom = n – 1 = 4

Sum of squares = X2 = df*S2 = 4*0.815 = 3.26

Sum of squares follows Chi square distribution. So, using Chi square table with value 3.26 and df = 4,

We get = 0.514295 or 51.4%

Part b

We are given

Population standard deviation = = 1

Population mean = µ = 3

We have to find P(X<1.9)

Z = (X - µ) /

Z = (1.9 – 3) / 1 = -1.1

P(Z<-1.1) = P(X<1.9) = 0.135666 (By using z-table or excel)

Required probability = 0.135666 or 13.57%

Part c

We have to find P(3<X<4)

P(3<X<4) = P(X<4) – P(X<3)

First we have to find P(X<4)

Z = (X - µ) /

We are given

Population standard deviation = = 1

Population mean = µ = 3

Z = (4 – 3) / 1 = 1

P(Z<1) = P(X<4) = 0.841345 (By using z-table or excel)

Now, we have to find P(X<3)

Z = (3 – 3) / 1 = 0

P(Z<0) = P(X<3) = 0.5000 (By using z-table or excel)

P(3<X<4) = P(X<4) – P(X<3)

P(3<X<4) = 0.841345 - 0.5000

P(3<X<4) =0.341345

Required probability = 0.341345 or 34.13%

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.