QuestionA light bulb lasts $N(1100, 60)$ hours. If 3/5 of bulbs are used, how many last between 1050 and 1140 hours?

Studdy Solution

STEP 1

Assumptions1. The lifespan of the light bulb follows a normal distribution. . The mean lifespan of the light bulb is1100 hours.
3. The standard deviation of the lifespan is60 hours.
4. We are interested in the proportion of light bulbs that last between1050 and1140 hours.

STEP 2

We need to convert the lifespan limits (1050 and1140 hours) into z-scores. The z-score is a measure of how many standard deviations an element is from the mean. The formula for the z-score is $z = \frac{X - \mu}{\sigma}$ where- X is the value we are interested in, - μ is the mean, - σ is the standard deviation.

STEP 3

First, let's calculate the z-score for the lower limit (1050 hours):
$z_{lower} = \frac{1050 -1100}{60}$

STEP 4

Calculate the z-score for the lower limit.
$z_{lower} = \frac{1050 -1100}{60} = -0.833$

STEP 5

Now, let's calculate the z-score for the upper limit (1140 hours):
$z_{upper} = \frac{1140 -1100}{60}$

STEP 6

Calculate the z-score for the upper limit.
$z_{upper} = \frac{1140 -1100}{60} =0.667$

STEP 7

Now we need to find the probability that a light bulb lasts between1050 and1140 hours. This is equivalent to finding the area under the normal distribution curve between the two z-scores we calculated. We can use a standard normal distribution table or a calculator with a normal distribution function to find these probabilities.
Let's denote the cumulative distribution function of the standard normal distribution as Φ(z). Then the probability we are looking for is $= Φ(z_{upper}) - Φ(z_{lower})$

STEP 8

Using a standard normal distribution table or a calculator, we find that $Φ(z_{lower}) = Φ(-0.833) =0.2023$ $Φ(z_{upper}) = Φ(0.667) =0.7475$

STEP 9

Now, substitute these values into the formula for $= Φ(z_{upper}) - Φ(z_{lower}) =.7475 -.2023$

STEP 10

Calculate the probability.
$=0.7475 -0.2023 =0.5452$

STEP 11

Now, we need to find the expected number of light bulbs that last between1050 and1140 hours. Since we know that the total number of light bulbs is3/5 of the total number of light bulbs in the building, we can multiply this by the probability we found to get the expected number.
Let's denote the total number of light bulbs in the building as N. Then the expected number of light bulbs that last between1050 and1140 hours is $Expected\, number = N \times \frac{3}{5} \times$

STEP 12

Without the actual total number of light bulbs (), we can't calculate the exact expected number. However, we can say that, to the nearest whole number, approximately54.52% of the/5 of the total light bulbs installed in the building are expected to last between1050 and1140 hours.

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