Question

12) On the basis of past experience, printers at a large South Dakota university computer lab are functioning 74% of the time on any given day. If a random sample of ten days are selected, what is the probability that printers are available: a) exactly eight days? b) all ten days? c) at most six days? d) more than seven days? e) fewer than six days? f) What is the standard deviation of the number of days for which the printers may be expected to be functioning? (Minitab usage is optional when answering this question.)

Answer 1

12)

Answer)

As there are fixed number of trials and probability of each and every trial is same and independent of each other

Here we need to use the binomial formula

P(r) = ncr*(p^r)*(1-p)^n-r

Ncr = n!/(r!*(n-r)!)

N! = N*n-1*n-2*n-3*n-4*n-5........till 1

For example 5! = 5*4*3*2*1

Special case is 0! = 1

P = probability of single trial = 0.74

N = number of trials = 10

R = desired success

A)

P(8) = 10c8*(0.74^8)*(1-0.74)^10-8 = 0.27353503997

B)

All 10

P(10) = 10c10*(0.74^10)*(1-0.74)^10-10 = 0.04923990397

C)

At most 6

P(0) + p(1) + p(2) + p(3) + p(4) + p(5) + p(6)

= 0.24793490555

D)

More than 7

P(8) + p(9) + p(10)

= 0.49578001196

E)

Less than 6

P(0) + p(1) + p(2) + p(3) + p(4) + p(5)

= 0.09035421294

F)

Standard deviation is = √{n*p*(1-p)}

N = 10

P = 0.74

S.d = 1.38708327075