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

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

Step-by-step 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

Answer 2

Answer:

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

Step-by-step 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 = $\frac{X-\mu}{\sigma}$ ~ 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( $\frac{X-\mu}{\sigma}$ <= $\frac{560-540}{20}$ ) = P(Z <= 1) = 0.8413

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