han want to buy a $30 ticket to a game, but the pre-oder tickets are sold out, he knowns there will be more tickets sold the day
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Answer:
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.
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