The company states that 90% of the light bulbs have a lifetime of at least k hours. Find the value of k. Give your answer correct to the nearest hundred.

a) Mean = 5800 hours

b) P(5000 ≤ L ≤ 6000) = 0.420

c) k = 4700 hours

Step-by-step explanation:

The full correct question is attached to this solution.

From the graph of the distribution of lifetime of bulbs presented, it is evident that this is a normal distribution.

(a) Write down the mean lifetime of the light bulbs.

From the normal distribution graph, the mean is at the very centre of the graph

Mean = μ = 5800 hours

(b) The standard deviation of the lifetime of the light bulbs is 850 hours.

Find the probability that 5000 ≤ L ≤ 6000, for a randomly chosen light bulb.

Mean = μ = 5800 hours

Standard deviation = σ = 850 hours.

The required probability = P(5000 ≤ L ≤ 6000)

To find this, we first normalize/standardize 5000 and 6000.

The standardized score for any value is the value minus the mean then divided by the standard deviation.

For 5000

z = (L- μ)/σ = (5000 - 5800)/850 = -0.941

For 6000

z = (L- μ)/σ = (6000 - 5800)/850 = 0.235

The required probability

P(5000 ≤ L ≤ 6000) = P(-0.941 ≤ z ≤ 0.235)

We'll use data from the normal probability table for these probabilities

P(5000 ≤ L ≤ 6000) = P(-0.941 ≤ z ≤ 0.235)

= P(z ≤ 0.235) - P(z ≤ -0.941)

= 0.593 - 0.173 = 0.420

c) The company states that 90% of the light bulbs have a lifetime of at least k hours. Find the value of k. Give your answer correct to the nearest hundred.

P(L ≥ k) = 90% = 0.9

assuming that the z-score of k is z'

P(L ≥ k) = P(z ≥ z') = 0.90

P(z ≥ z') = 1 - P(z < z') = 0.9

P(z < z') = 0.1

Using the normal distribution table

z' = -1.282

z' = (k - μ)/σ

-1.282 = (k - 5800)/850

k = (-1.282 × 850) + 5800 = 4710.3 = 4700 hours to the nearest hundred.

Hope this Helps!!!

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