- #1
~angel~
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Note: You'll need the Normal Distribution Table.
A certain type of light bulb has a lifetime in hours which is normally distributed with mean μ=650 and standard deviation σ=40. What is the probablility that a randomly selected light bulb has a lifetime in the range (700, 850)?
Now this is how I tried to do this:
We have to find P(700 ≤ X ≤ 850), but we need it in terms of Z.
Now,
Z= (X-μ)/σ
Z= (X-650)/40
But we have to apply this to all the sides of the inequality, so we end up with
P(1.25 ≤ (X-650)/40 ≤ 5)
= P(1.25 ≤ Z ≤ 5)
But using the Normal Distribution Table, there is no probability for the "5" bit of the inequality. So I'm not sure how to do this.
Thanks.
A certain type of light bulb has a lifetime in hours which is normally distributed with mean μ=650 and standard deviation σ=40. What is the probablility that a randomly selected light bulb has a lifetime in the range (700, 850)?
Now this is how I tried to do this:
We have to find P(700 ≤ X ≤ 850), but we need it in terms of Z.
Now,
Z= (X-μ)/σ
Z= (X-650)/40
But we have to apply this to all the sides of the inequality, so we end up with
P(1.25 ≤ (X-650)/40 ≤ 5)
= P(1.25 ≤ Z ≤ 5)
But using the Normal Distribution Table, there is no probability for the "5" bit of the inequality. So I'm not sure how to do this.
Thanks.