Question

(Use computer) Let X represent a binomial random variable with n = 180 and p = 0.23. Find the following probabilities. (Round your final answers to 4 decimal places.) a. P(X ≤ 45) b. P(X = 35) c. P(X > 55) d. P(X ≥ 50)

Answer 1

X represent a binomial random variable with n = 180 and p = 0.23, so p = 1- p = 1 - 0.23 = 0.77

n*p = 180*0.23 = 41.4 and n*q = 180*0.77 = 138.6

Since both n*p and n*q are greater than 5 , we have to use normal approximation to find the all given probabilities.

First we need to find Mean and Standard deviation (SD)

Mean =n*p = 41.4

SD = $\sqrt{n*p*q}$ = $\sqrt{180*0.23*0.77}$ = 5.6461

Before using the normal approximation , first we need to use continuity correction for X

The following table gives the continuity rules for different probability equations.

a)P(X ≤ 45)  

Therefore according to above continuity correction table we need to add 0.5 to 45

P(X ≤ 45.5)

We can find the normal probability using excel function =NORM.DIST( x , Mean , SD, cumulative )

For x = 45.5 , then we plug the calculated value of mean and SD accordingly , for cumulative plug "TRUE"

=NORM.DIST(45.5, 41.4,5.6461,TRUE)

P(X ≤ 45.5)  =0.7661

b) P(X = 35)

We need to add and subtract 0.5 from 35

P( 34.5 <= x <= 35.5) = P( x <= 35.5 ) - P( x <= 34.5)

For P( x <= 35.5 ) type =NORM.DIST(35.5, 41.4,5.6461,TRUE) in the excel.

P( x <= 35.5 ) = 0.1480

For P( x <= 34.5) type =NORM.DIST(34.5, 41.4,5.6461,TRUE) in the excel.

P( x <= 34.5) = 0.1108

P( 34.5 < x < 35.5) = P( x <= 35.5 ) - P( x <= 34.5) = 0.1480 - 0.1108 = 0.0372

P(X = 35) = 0.0372

c. P(X > 55)

We need to add 0.5 to 55

P(X => 55.5)

type =1 - NORM.DIST(55.5, 41.4,5.6461,TRUE) in the excel.

P(X > 55) = 0.0063

d. P(X ≥ 50)

We need to subtract 0.5 from 50

P(X ≥ 49.5)

type =1 - NORM.DIST(49.5, 41.4, 5.6461,TRUE) in the excel.

P(X ≥ 50) = 0.0757