Gears produced by a grinding process are categorized

Solution 14E

Step1 of 5:

We have random variable X which presents the number of gears that are scrap among the ten selected. Here X follows binomial distribution with parameters “n and p” that is X B(n, p),

The probability mass function of binomial distribution is given by

, x = 0,1,2,...,n.

Where,

n = sample size

   = 10

x = random variable

p = probability of success

   = 5%

   = 0.05.

q = 1 - p (probability of failure)

   = 1 - 0.05

   = 0.95

Here our goal is:

a).We need to find the probability that one or more is scrap.

b).We need to find the probability that eight or more are not scrap.

c).We need to find the probability that more than two are either degraded or scrap.

d).We need to find the probability that exactly nine are either conforming or degraded.

Step2 of 5:

a).

P(One or more is scrap) = P(X1)

                                        = 1 - P(X < 1)

                                        = 1 - P(X = 0)

Consider,

       P(X) =

P(X = 0) =  

             = (1)(1)(0.5987)

             = 0.5987

Now,

P(one or more is scrap) = 1 - P(X = 0)

                                      = 1 - 0.5987

                                      = 0.4012

Hence, P(one or more is scrap) = 0.4012.

Step3 of 5:

b).

We have total gears n = 10 in that we need to find the probability that eight or more are not scrap

Hence,

P(eight or more are not scrap) = P(two or more are scrap)

                                           = P(X2)

1).Mean of the binomial distribution is

                                                            =

                                                            = 0.5

     Hence, = 0.5.  

2).Standard deviation of binomial distribution is

                                                                        =

                                                                        =  

                                                                        = 0.6892

    Hence, =0.6892.

 Now,

P(X2) =

              =

              =

value is obtained from standard normal table(area under normal curve) then

= 0.9850 (In area under normal curve we have to see row 2.1 under column 0.07)

 Therefore,  0.9850.

Step4 of 5:

c).

From the given information we have 15% degraded and 5% are Scrap and total selected gears

n = 10.

That is p = 0.15 + 0.05

               = 0.20

P(more than two are either degraded or scrap) = P(X)

                                                                = 1 - P(X2)

                                                                = 1 - {P(X=0)+P(X=1)+P(X=2)}

Consider,

{P(X=0)+P(X=1)+P(X=2)} =

                                             = {1(1)(0.1073)+10(0.2)(0.1342)+45(0.04)(0.1677)}

                                             = {0.1073+0.2684+0.3018}

                                             = 0.6775

Now,

P(X) = 1 - {P(X=0)+P(X=1)+P(X=2)}

               = 1 - 0.6775

               = 0.3224

Hence, P(X) = 0.3224.

Step5 of 5:

d).

We have n = 10 and p = 0.95

Consider,

P(exactly nine are either conforming or degraded) = P(X = 9)

                                                                      =

                                                                      = 10(0.6302)(0.05)

                                                                      = 0.3151

Hence, P(X = 9) = 0.3151.