When circuit boards used in the manufacture of compact disc

When circuit boards used in the manufacture of compact disc players are tested, the long-run percentage of defectives is10%. Let X = the number of defective boards in a random sample of size n = 25, so X ~ Bin(25, 0.1). (Round your probabilities to three decimal places.)

(a) Determine P(X 2).


(b) Determine P(X 5).


(c) Determine P(1 X 4).


(d) What is the probability that none of the 25 boards is defective?


(e) Calculate the expected value and standard deviation of X. (Round your standard deviation to two decimal places.)

You may need to use the appropriate table in the Appendix of Tables to answer this question.

Solution

a)

Using a cumulative binomial distribution table or technology, matching          
          
n = number of trials =    25      
p = the probability of a success =    0.1      
x = the maximum number of successes =    2      
          
Then the cumulative probability is          
          
P(at most   2   ) =    0.53709405

b)

Note that P(at least x) = 1 - P(at most x - 1).          
          
Using a cumulative binomial distribution table or technology, matching          
          
n = number of trials =    25      
p = the probability of a success =    0.1      
x = our critical value of successes =    5      
          
Then the cumulative probability of P(at most x - 1) from a table/technology is          
          
P(at most   4   ) =    0.902006379
          
Thus, the probability of at least   5   successes is  
          
P(at least   5   ) =    0.097993621

c)

Note that P(between x1 and x2) = P(at most x2) - P(at most x1 - 1)          
          
Here,          
          
x1 =    1      
x2 =    4      
          
Using a cumulative binomial distribution table or technology, matching          
          
n = number of trials =    25      
p = the probability of a success =    0.1      
          
Then          
          
P(at most    0   ) =    0.071789799
P(at most    4   ) =    0.902006379
          
Thus,          
          
P(between x1 and x2) =    0.83021658      

d)

Note that the probability of x successes out of n trials is          
          
P(n, x) = nCx p^x (1 - p)^(n - x)          
          
where          
          
n = number of trials =    25      
p = the probability of a success =    0.1      
x = the number of successes =    0      
          
Thus, the probability is          
          
P (    0   ) =    0.071789799

e)

E(x) = n p = 25*0.1 = 2.5
s(x) = sqrt(n p (1-p)) = sqrt(25*0.1*(1-0.1)) = 1.5 [answer]