Question

For n = 80 and 1 = 0.6, use the normal distribution to approximate the following probabilities. a. X= 50 b. X> 50 c. X 550 d. X<50 a. The approximate probability that X = 50 is 17. (Round to four decimal places as needed.)

Answer 1

Answer)

N = 80

P = 0.6

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 = 48

N*(1-p) = 32

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

As the data is normally distributed we can use standard normal z table to estimate the answers

Z = (x-mean)/s.d

Given mean = n*p = 48

S.d = √{n*p*(1-p)} = 4.38178046004

A)

P(x=50) = p(49.5<x<50.5) = p(x<50.5) - p(x<49.5)

P(x<50.5)

Z = (50.5 - 48)/4.38178046004 = 0.57

From z table, P(z<0.57) = 0.7157

P(x<49.5)

Z = 0.34

From z table, P(z<0.34) = 0.6331

Required probability is 0.7157 - 0.6331

= 0.0826

B)

P(x>50)

By continuity correction

P(x>50.5)

Z = 0.57 (calculated in part a)

From z table p(z>0.57) = 0.2843

C)

P(x<=50)

By continuity correction

P(x<50.5)

Z = 0.57

From z table, p(z<0.57) = 0.7157

D)

P(x<50)

By continuity correction

P(x<49.5)

Z = 0.34

From z table, P(z<0.34) = 0.6331