University:
California State University, Northridge
Course:
MATH 262 | Introduction to Linear Algebra
Academic year:

2024
Views:

155

Pages:

3
Author:

Beckett Gray

Question 1 Use the normal approximation to the binomial distribution to determine (to four decimals) the probability of getting 7 heads and 7 tails in 14 flips of a balanced coin. To determine the error of this approximation, also consult the 'Statistical Tables' table of binomial probabilities. Answer Step 1 According to the information given, the event involves 14 balanced coin flips. 𝑛 = 14 n=14. As flipping a coin has two outcomes as head or tail, the probability of getting head or tail is 0.5. The probability mass function of binomial distribution is, 𝑃 ( 𝑋 = 𝑥 ) = 𝐶 𝑛 𝑥 𝑝 𝑥 ( 1 − 𝑝 ) 𝑛 − 𝑥 P(X=x)=C n x p x (1−p) n−x With mean 𝑛 𝑝 = 7 np=7 and variance 𝑛 𝑝 ( 1 − 𝑝 ) = 3.5 np(1−p)=3.5. For normal approximation 𝑋 ∼ 𝑁 ( 𝑛 𝑝 , 𝑛 𝑝 ( 1 − 𝑝 ) ) X∼N(np,np(1−p)). Step 2 The probability that there are exactly 7 heads can be calculated as: 𝑃 ( 𝑋 = 7 ) = 𝑃 ( 7 − 0.5 < 𝑋 < 7 + 0.5 ) P(X=7)=P(7−0.5 13 ) = 𝑃 ( 𝑋 > 13.5 ) P(X>13)=P(X>13.5) = 1 − 𝑃 ( 𝑋 < 13.5 ) =1−P(X<13.5) = 1 − 𝑃 ( 𝑋 − 𝑛 𝑝 𝑛 𝑝 𝑞 < 13.5 − ( 20 ⋅ 0.8 ) 20 ⋅ 0.8 ⋅ 0.2 ) =1−P( npq X−np < 20⋅0.8⋅0.2 13.5−(20⋅0.8) ) = 1 − 𝑃 ( 𝑍 < − 1.39754 ) =1−P(Z<−1.39754) = 1 − 0.081125 =1−0.081125 = 0.918875 =0.918875 Since the area corresponding to 𝑍 = − 1.39754 Z=−1.39754 is 0.081125 0.081125, calculated by Excel command NORM.S.DIST(-1.39754, TRUE). Thus, the probability is 0.919. Question 4 Compute the probability of 𝑥 x successes in the 𝑛 n independent trials of the binomial experiment. 𝑛 = 10 , 𝑝 = 0.6 , 𝑥 = 7 n=10,p=0.6,x=7 𝑃 ( 𝑋 = 7 ) ? P(X=7)? Answer Step 1 Given Information: Number of trials 𝑛 = 10 n=10 Probability of success 𝑝 = 0.6 p=0.6. To compute the probability of 𝑋 = 7 X=7 successes: Binomial formula: 𝑃 ( 𝑋 = 𝑥 ) = 𝐶 𝑛 𝑥 𝑝 𝑥 ( 1 − 𝑝 ) 𝑛 − 𝑥 P(X=x)=C n x p x (1−p) n−x Step 2 Required probability is obtained as follows: 𝑃 ( 𝑋 = 7 ) = 𝐶 10 7 ( 0.6 ) 7 ( 1 − 0.6 ) 10 − 7 P(X=7)=C 10 7 (0.6) 7 (1−0.6) 10−7 = 10 ! 7 ! ( 10 − 7 ) ! ( 0.6 ) 7 ( 0.4 ) 3 = 7!(10−7)! 10! (0.6) 7 (0.4) 3 = 10 ⋅ 9 ⋅ 8 3 ⋅ 2 ⋅ 1 ( 0.6 ) 7 ( 0.4 ) 3 = 3⋅2⋅1 10⋅9⋅8 (0.6) 7 (0.4) 3 = 120 ⋅ 0.0279936 ⋅ 0.064 =120⋅0.0279936⋅0.064 = 0.2150 =0.2150 Therefore, required probability is 0.2150. Question 5 To define: The continuity correction. Explanation: The procedure for applying the continuity correction method for finding the probability if using the normal approximation to the binomial distribution is used for finding the probability. Given info: Out of 14 couples, 13 of them had baby girls by using the preliminary test of the MicroSoft method of gender selection. Answer Calculation: Continuity correction: The continuity correction is used to approximate the binomial distribution (discrete) by using the normal distribution (continuous). If the binomial probability represents “exactly” then the value 0.5 is added and subtracted with the value of 𝑥 x. The couples having exactly 13 girls.