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.

n=54 p=0.4 and x= 17

For

nequals=5454​,

pequals=0.40.4​,

and

Xequals=1717​,

use the binomial probability formula to find​ P(X).

0.05010.0501

​(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 10np(1−p)≤10

B.

​Yes, because np left parenthesis 1 minus p right parenthesis greater than or equals 10np(1−p)≥10

C.

​Yes, because StartRoot np left parenthesis 1 minus p right parenthesis EndRoot greater than or equals 10np(1−p)≥10

D.

​No, because np left parenthesis 1 minus p right parenthesis less than or equals 10np(1−p)≤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

Answer:

Given,

sample n = 54 , x = 17 , p = 0.4

Consider,

Binomial distribution P(x) = nCx*p^x*q^(n-x)

P(X = 17) = 54C17*0.4^17*(1-0.4)^(54-17)

= 54C17*0.4^17*0.6^37

= 0.0501

np(1-p) = 54*0.4(1-0.4) = 12.96 >= 10

Yes, the normal distribution can be used as p(1-p) >= 10

Mean = np = 54*0.4 = 21.6

variance = npq = 12.96

standard deviation = sqrt(npq) = 3.6

P(X) = N(21.6 , 12.96)

Here f(x) = 1/[*sqrt(2) ] *e^-(x-u)^2/2^2

x = 17

substitute values & then we get

= 0.0490

Exact & approximated difference = 0.0501 - 0.0490

= 0.0011