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
(Interchange the x and y-values)
(Solve for y)
