Question

Assume that on a standardized test of 100 questions, a person has a probability of 80% of answering any particular question correctly. Find the probability of answering between 70 and 80 questions, inclusive. (Assume independence, and round your answer to four decimal places.)

P(70 ≤ X ≤ 80) =

Answer 1

Solution:

Given that,

P = 80% = 0.80

1 - P = 1 - 0.80 = 0.20

n = 100

Here, BIN ( n , P ) that is , BIN (100 , 0.80)

then,

n*p = 100* 0.80 = 80 > 5

n(1- P) = 100*0.20 = 20 > 5

According to normal approximation binomial,

X Normal

Mean = = n*P = 80

Standard deviation = =n*p*(1-p) = 100*0.80*0.20= 16 = 4

We using countinuity correction factor

P(a-0.5 Xb + 0.5 ) = P( 70 - 0.5 < X < 80 + 0.5)

= P((69.5 80\4<(x - ) /  < (80.5-80)/4))

= P( -2.625 < z < 0.125)

= P(Z < 0.125) - P(Z < -2.625)

= 0.5497 - 0.0043

= 0.5454

Probability = 0.5454