Question

Ten percent of the engines manufactured on an assembly line are defective.

Answer 1

Answer:

a) the probability that the first non-defective engine will be found on the second trial is 0.09

b) probability that the third non-defective engine will be found on the fifth trial is 0.00486

c) the Mean is 1.1111  and Variance is 0.1235  

d) the Mean is 3.3333and Variance is 0.3704  

Step-by-step explanation:

Firstly;

Let x be the number of trail on which the rth defective occurs

Also let the probability that an occurrence of a defective be p = 0.10

Here, x follows negative binomial distribution with parameters r and p

The probability mass function of X is as follows:

P(X = x) = [ x -1  p^r ( 1 - p)^(x-r)       ; x = r, r + 1, r + 2, ...

                 r - 1 ]

=   [ x -1  (0.10)^r ( 1 - 0.10)^(x-r)

     r - 1 ]

= [ x -1  (0.10)^r ( 0.9)^(x-r)  ..........n 1..... let this be equatio

    r - 1 ]

This represents the probability that the rth success occurs on the xth trail.  

a)

probability that the first non-defective engine will be found on the second trial?

Substitute r = 1 and x = 2 in equation 1  

P(X = 2)  = [ 2 - 1   (0.10)¹ ( 0.90 )²⁻¹

                   1 - 1 ]

= (0.10) × (0.90)

= 0.09

Therefore, the probability that the first non-defective engine will be found on the second trial is 0.09

b)

probability that the third non-defective engine will be found on the fifth trial?

So we substitute r = 3 and x = 5 in equation 1  

P(X = 5) = [ 5 - 1  (0.10)³ ( 0.90)⁵⁻³

                  3 - 1 ]

=    [ 4    (0.10)³ ( 0.9)²

      2 ]

= 0.00486

Therefore, probability that the third non-defective engine will be found on the fifth trial is 0.00486

Now the formula for the mean of the negative binomial distribution is as follows:  

Mean u = r / p    ------- let this be equation 2

The formula for the variance of the negative binomial distribution also is as follows:  

Variance α² = rq / p²   ---------- let this be equation 3

so

c)

the mean and variance of the number of trials on which the first non-defective engine is found.  

First, let the probability that non-defective engine found be p = 0.90

And q = (1 - p) = 1 - 0.90 = 0.10  

Now we substitute r = 1, p = 0.90 and q = 0.10 in equation 2 & 3simultaneously,

the mean and variances are as follows;

Mean = r/p = 1/0.90 = 1.1111

Variance = rq/p² = (1)(0.10) / (0.90)² = 0.1235  

Therefore the Mean is 1.1111  and Variance is 0.1235  

d)

the mean and variance of the number of trials on which the third non-defective engine is found  

Let the probability that non-defective engine found be p = 0.90

And q = (1 - p) = 1 - 0.90 = 0.10

Now we substitute r = 3, p = 0.90 and q = 0.10 in equation (2) and (3) simultaneously,

the mean and variances are;

Mean = r/p = 3/0.90 = 3.3333

Variance = rq/p² = (3)(0.10) / (0.90)² = 0.3704

Therefore the Mean is 3.3333 and Variance is 0.3704