Question

A new extended-life light bulb has an average life of 750 hours, with a standard deviation of 50 hours. If the life of these light bulbs approximates a normal distribution, about what percent of the distribution will be between 650 hours and 850 hours?

A. 95%
B. 68%
C. 99%
D. 34%

Answer 1

95% of the distribution will be between 650 hours and 850 hours.

Explanation:

We are given that a new extended-life light bulb has an average life of 750 hours, with a standard deviation of 50 hours.

The life of these light bulbs approximates a normal distribution.

Let X = life of these light bulbs

So, X ~ Normal(
`\mu=750,\sigma^(2) =50^(2)`)

The z-score probability distribution for normal distribution is given by;

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

where,
`\mu` = average life of light bulb = 750 hours

`\sigma` = standard deviation = 50 hours

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, Probability that the distribution will be between 650 hours and 850 hours is given by = P(650 hours < X < 850 hours)

P(650 hours < X < 850 hours) = P(X < 850 hours) - P(X
`\leq` 650 hours)

P(X < 850 hours) = P(
`( X-\mu)/(\sigma)} }` <
`( 850-750)/(50)} }` ) = P(Z < 2) = 0.97725

P(X
`\leq` 650 hours) = P(
`( X-\mu)/(\sigma)} }`
`\leq`
`( 650-750)/(50)} }` ) = P(Z
`\leq` -2) = 1 - P(Z < 2)

= 1 - 0.97725 = 0.02275

So, in the z table the P(Z
`\leq` x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 2 in the z table which has an area of 0.97725.

Therefore, P(650 hours < X < 850 hours) = 0.97725 - 0.02275 = 0.9545 or 95%.

Hence, 95% of the distribution will be between 650 hours and 850 hours.