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 no

Question

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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Melody 3 months 2021-10-20T07:02:23+00:00 1 Answer 0 views 0

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    2021-10-20T07:04:15+00:00

    Answer:

    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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