Question

Compute​ P(X) using the binomial probability formula. Then determine whether the normal distribution can be used to estimate this probability. If​ so, approximate​ P(X) using the normal distribution and compare the result with the exact probability. nequals44​, pequals0.4​, and Xequals19 For nequals44​, pequals0.4​, and Xequals19​, use the binomial probability formula to find​ P(X). nothing ​(Round to four decimal places as​ needed.) Can the normal distribution be used to approximate this​ probability? A. ​No, because StartRoot np left parenthesis 1 minus p right parenthesis EndRoot less than or equals 10 B. ​No, because np left parenthesis 1 minus p right parenthesis less than or equals 10 C. ​Yes, because np left parenthesis 1 minus p right parenthesis greater than or equals 10 D. ​Yes, because StartRoot np left parenthesis 1 minus p right parenthesis EndRoot greater than or equals 10 Approximate​ P(X) using the normal distribution. Use a standard normal distribution table. A. ​P(X)equals nothing ​(Round to four decimal places as​ needed.) B. The normal distribution cannot be used. By how much do the exact and approximated probabilities​ differ? A. nothing ​(Round to four decimal places as​ needed.) B. The normal distribution cannot be used

Answer 1

n = 44

p = 0.4

P(X = x) = nCx * p^x * (1 - p)^{n - x}

P(X = 19) = 44C19 * (0.4)^19 * (0.6)^25 = 0.1101

np(1 - p) = 44 * 0.4 * (1 - 0.4) = 10.56

Since np(1 - p) > 10, so we can use normal approximation to the binomial distribution.

Option - C) Yes, because np(1 - p) > 10.

= np = 44 * 0.4 = 17.6

= sqrt(np(1 - p))

= sqrt(44 * 0.4 * (1 - 0.4))

= 3.2496

P(X = 19)

= P((18.5 - )/< (X - )/< (19.5 - )/)

= P((18.5 - 17.6)/3.2496 < Z < (19.5 - 17.6)/3.2496)

= P(0.28 < Z < 0.58)

= P(Z < 0.58) - P(Z < 0.28)

= 0.7190 - 0.6103

= 0.1087