Question

Consider a binomial experiment with 15 trials and probability 0.55 of success on a single trial.

(a) Use the binomial distribution to find the probability of exactly 10 successes. (Round your answer to three decimal places.)


(b) Use the normal distribution to approximate the probability of exactly 10 successes. (Round your answer to three decimal places.)


(c) Compare the results of parts (a) and (b).

These results are fairly different.These results are almost exactly the same.  

Answer 1

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 = 0.55

N = number of trials = 15

R = desired success = 10

P(10) = 15c10*(0.55^10)*(1-0.55)^15-10 = 0.14036046426

B)

N = 15

P = 0.55

First we need to check the conditions of normality that is if n*p and n*(1-p) both are greater than 5 or not

N*p = 15*0.55 = 8.25

N*(1-p) = 6.75

Both the conditions are met so we can use standard normal z table to estimate the probability

Z = (x - mean)/(s.d)

Mean = n*p = 8.25

S.d = √{n*p*(1-p)} = 1.92678488679

P(10) = P(9.5<x<10.5) = P(x<10.5) - P(x<9.5)

P(x<10.5)

Z = (10.5-8.25)/(1.92678488679) = 1.17

From z table, P(z<1.17) = 0.879

P(x<9.5)

Z = 0.65

From z table, P(z<0.65) = 0.7422

Probability = 0.879 - 0.7422 = 0.1368

C)

Probability in part b is less than part a