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forschoolok123456789
04.03.2020 •
Mathematics
Suppose that a brand of lightbulb lasts on average 1730 hours with a standard deviation of 257 hours. Assume the life of the lightbulb is normally distributed. Calculate the probability that a particular bulb will last from 1689 to 2267 hours?
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Ответ:
P [ 1689 ≤ X ≤ 2267 ] = 54,88 %
Step-by-step explanation:
Normal Distribution
Mean μ₀ = 1730
Standard Deviation σ = 257
We need to calculate z scores for the values 1689 and 2267
We apply formula for z scores
z = ( X - μ₀ ) /σ
X = 1689 then
z = (1689 - 1730)/ 257 ⇒ z = - 41 / 257
z = - 0.1595
And from z table we get for z = - 0,1595
We have to interpolate
- 0,15 0,4364
- 0,16 0,4325
Δ = 0.01 0.0039
0,1595 - 0,15 = 0.0095
By rule of three
0,01 0,0039
0,0095 x ?? x = 0.0037
And 0,4364 - 0.0037 = 0,4327
Then P [ X ≤ 1689 ] = 0.4327 or P [ X ≤ 1689 ] = 43,27 %
And for the upper limit 2267 z score will be
z = ( X - 1730 ) / 257 ⇒ z = 537 / 257
z = 2.0894
Now from z table we find for score 2.0894
We interpolate and assume 0.9815
P [ X ≤ 2267 ] = 0,9815
Ths vale already contains th value of P [ X ≤ 1689 ] = 0.4327
Then we subtract to get 0,9815 - 0,4327 = 0,5488
Finally
P [ 1689 ≤ X ≤ 2267 ] = 0,5488 or P [ 1689 ≤ X ≤ 2267 ] = 54,88 %
Ответ:
can you repost with a clearer picture of the numbers?