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    1. Find the value of (a) in the following discrete probability distribution:
    X -2 0 2
    P(x): a 0.35 0.25

    2. A binomial distribution is based on n=25 and p=0.1. Find the probability that x =1.

    3 A production process produces parts with weights that are normally distributed with a mean of 1.75 ounces and a standard deviation of 0.15 ounces. If specifications call for weights between 1.50 and 2.25 ounces, what percent are not within specifications?

    4 If the scores on a test were normally distributed with a mean equal to 500 and a standard deviation equal to 100, what test score would produce a z score equal to 1.5?

    5 A certain intersection has a stop light that some residents feel is inappropriately placed. They estimate that 25% of all cars through the intersection do not stop for a red light. If this is true, how many in a sample of 300 cars encountering a red light would one expect to see not stop? What is the probability that less than 20% of the 300 cars would be observed running a red light?

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    https://brainmass.com/statistics/probability/distributions-discrete-probability-distribution-6943

    SOLUTION This solution is FREE courtesy of BrainMass!

    1. Sum of Probabilities should be 1.
    So a + .35+ .25 = 1
    or a = .4

    2. P(X=1) = (n C x)*(p^x)*(1-p)^n-x
    = 25C1* .1^1*.9^24
    = .199

    3. Mean(M) = 1.75
    Std Dev = .15
    Lower Limit = 1.5
    Upper Limit = 2.25

    Z lower = (1.5 - 1.75)/.15 ( Z is a normal variate given by Z = (X- Mean)/(Std Dev)...here X is our limits.

    Z lower = -1.66

    Similarly Z upper = (2.25 - 1.75)/.15
    = 3.333

    So P(1.5<=X<=2.25) = P(-1.66<=Z<=3.33)
    From Normal tables we have P= .9511 ( Not this P(Z<=3.33) - P(Z<=-1.66)..so .999 - .0484)

    Percent no within specs = 1- .9511 = .0488

    4. M= 500
    Std Dev = 100
    z = 1.5
    z = ( x-m)/(std dev)
    so 1.5 = (x - 500)/100
    so X = 650.

    5. p = .25
    n = 300
    So expected no of cars not stopping = n*p
    = 300*.25 = 75
    Prob that less than 20% cars , so Prob that that less than 60 cars( 20% of 300) jump the light.
    Poisson Distribution P ( X<=60) = .0433 ( Note this is claculated using an inbuilt function called POISSON in Excel.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com October 2, 2022, 6:55 pm ad1c9bdddf>
    https://brainmass.com/statistics/probability/distributions-discrete-probability-distribution-6943

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