Question

The life of light bulbs is distributed normally. The variance of the lifetime is 400 and the mean lifetime of a bulb is 540 hours. Find the probability of a bulb lasting for at most 560 hours. Round your answer to four decimal places.

Answer 1

Probability of a bulb lasting for at most 560 hours = 0.8413 .

Explanation:

We are given that the life of light bulbs is distributed normally. The variance of the lifetime is 400 and the mean lifetime of a bulb is 540 hours.

Let X = life of light bulbs

So, X ~ N(
`\mu = 540,\sigma^(2) =20^(2)`)

The z score probability distribution is given by;

Z =
`(X-\mu)/(\sigma)` ~ N(0,1)

where,
`\mu` = population mean and
`\sigma` = standard deviation

So, Probability of a bulb lasting for at most 560 hours = P(X <= 560 hours)

P(X <= 560) = P(
`(X-\mu)/(\sigma)` <=
`(560-540)/(20)` ) = P(Z <= 1) = 0.8413

Therefore, probability of a bulb lasting for at most 560 hours is 0.8413 .

Answer 2

Answer: The probability of a bulb lasting for at most 560 hours is 0.84

Explanation:

Since the life of light bulbs is distributed normally, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = life of light bulbs.

µ = mean lifetime

σ = standard deviation

From the information given,

µ = 540 hours

Variance = 400

σ = √variance = √400

σ = 20

The probability of a bulb lasting for at most 560 hours is expressed as

P(x ≤ 560)

For x = 560

z = (560 - 540)/20 = 1

Looking at the normal distribution table, the probability corresponding to the z score is 0.84