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Math  /  Data & Statistics

Question19. Battery Life The length of life (in days) of an alkaline battery has an exponential distribution with an average life of 1 year, so that λ=1/365\lambda=1 / 365. a. What is the probability that an alkaline battery will fail before 180 days?
Answer b. What is the probability that an alkaline battery will last beyond 1 year?
Answer c. If a device requires two batteries, what is the probability that both batteries last beyond 1 year?

Studdy Solution

STEP 1

1. The battery life follows an exponential distribution with a rate parameter λ=1365\lambda = \frac{1}{365}.
2. The exponential distribution probability density function is given by f(x;λ)=λe−λx f(x; \lambda) = \lambda e^{-\lambda x} .
3. The cumulative distribution function (CDF) for an exponential distribution is F(x;λ)=1−e−λx F(x; \lambda) = 1 - e^{-\lambda x} .
4. The probability that a battery lasts beyond a certain time is the complement of the CDF.

STEP 2

1. Calculate the probability that a battery will fail before 180 days.
2. Calculate the probability that a battery will last beyond 1 year.
3. Calculate the probability that both batteries in a device last beyond 1 year.

STEP 3

Calculate the probability that a battery will fail before 180 days using the CDF:
F(180;λ)=1−e−1365×180 F(180; \lambda) = 1 - e^{-\frac{1}{365} \times 180}

STEP 4

Compute the value:
F(180;λ)=1−e−180365 F(180; \lambda) = 1 - e^{-\frac{180}{365}}
F(180;λ)=1−e−0.49315 F(180; \lambda) = 1 - e^{-0.49315}
F(180;λ)≈1−0.6107 F(180; \lambda) \approx 1 - 0.6107
F(180;λ)≈0.3893 F(180; \lambda) \approx 0.3893
The probability that an alkaline battery will fail before 180 days is approximately 0.3893 \boxed{0.3893} .

STEP 5

Calculate the probability that a battery will last beyond 1 year using the complement of the CDF:
P(X>365)=1−F(365;λ)=e−1365×365 P(X > 365) = 1 - F(365; \lambda) = e^{-\frac{1}{365} \times 365}

STEP 6

Compute the value:
P(X>365)=e−1 P(X > 365) = e^{-1}
P(X>365)≈0.3679 P(X > 365) \approx 0.3679
The probability that an alkaline battery will last beyond 1 year is approximately 0.3679 \boxed{0.3679} .

STEP 7

Calculate the probability that both batteries last beyond 1 year. Since the batteries are independent, the probability is the product of the individual probabilities:
P(Both last>365)=P(X1>365)×P(X2>365) P(\text{Both last} > 365) = P(X_1 > 365) \times P(X_2 > 365)
P(Both last>365)=(0.3679)2 P(\text{Both last} > 365) = (0.3679)^2

STEP 8

Compute the value:
P(Both last>365)=0.3679×0.3679 P(\text{Both last} > 365) = 0.3679 \times 0.3679
P(Both last>365)≈0.1353 P(\text{Both last} > 365) \approx 0.1353
The probability that both batteries last beyond 1 year is approximately 0.1353 \boxed{0.1353} .

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