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17 August, 02:04

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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  1. 17 August, 02:30
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    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 %
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