Question

Consider a binomial experiment with 20 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

solution:-

given that n = 20 , p = 0.55

(a) by using binomial distribution

formula P(X = x) = ncx * p^x * (1-p)^(n-x)

=> P(X = 10) = 20c10 * 0.55^10 * (1-0.55)^(20-10)

= 20!/(10!(20−10)!) * 0.55^10 * (1-0.55)^(20-10)

= 184756 * 0.55^10 * (1-0.55)^(20-10)

= 0.159

(b) by using normal distribution

mean = n*p = 20*0.55 = 11

standard deviation = sqrt(n*p*(1-p)) = sqrt(20*0.55*0.45) = 0.2249

=> P(X = 10)

=> P(9.5 < x < 10.5)

=> P((9.5-11)/2.2249 < z < (10.5-11)/2.2249)

=> P(-0.67 < z < -0.22)

=> 0.162

(c) compare the results of part (a) and (b)

=> option: these results are almost exactly the same