Statistics and Probabilities.. Send help please?

Suppose family incomes in a town are normally distributed with a mean of P25000

and a standard deviation of P6000 per month. What is the probability that a family

has an income between P14,000 and P32,500? 

2 Answers

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  • Alan
    Lv 7
    4 weeks ago

    P( 14000< X < 32500) = P(x<32500) - P(x<14000) 

    P(x< 14000) = P( Z < (14000-25000)/6000 )  = P(z< -1.8333) 

    P(x< 14000) = P(z< -1.833333)

    Using this online z-table, 

    https://www.math.arizona.edu/~rsims/ma464/standard... 

    P(z< -1.84) = 0.03288 

    P(z< -1.83) = 0.03362

    Z = -1.83333 

    Interpolating 

    p = 0.03288 + (-1.8333333 - (-1.84))*(0.03362-0.03288)/ 0.01 

    p(x<14000)  = 0.03337

    p(x<32500) = P( z< (32500-25000)/6000 ) = P(z< 1.25) 

    P(x<32500) =  0.89435 

     P( 14000< X < 32500) =  0.89435 - 0.03337 =  0.86098

    P( 14000< X < 32500) = 0.86098

    (round as you wish) 

  • rotchm
    Lv 7
    4 weeks ago

    Transform your ranges into z space via z = (x-avg)/sd.

    Once you have your z's, lookup the corresponding area in your z table.

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