In: Statistics and Probability
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) =
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