Question

Find to 4 decimal places the following binomial probabilities using the normal approximation. a. n-150, p0.44, P(x-73) P(x-73)- b. n-100, ?-0.56, P(52 x 59) P(52 Sx s 59)- P(x 2 40)- d. n-105, ?-0.72, P(x 78) Click if you would like to Show Work for this question: Open Show Work

Answer 1

a) X ~ Binornial(n-150, p=0.44) Binomial can be approximated to normal with: σ = v/np(1-p) = V 150 * (0.44) (1-0.44) 6.079474 6.079 P(X = 73) Since we are approximating a discrete distribution by continuous normal distri bution, values between 72.5 and 73.5 both approrimate to 73. Thus, "equal to 73 "corresponds to continuous normal distribution with P(72.5 X< 73.5) after continuity correction Normal Distribution, zi-72.5, x2-73.5, μ-66, σ-6.079 We convert this to standard normal using z- 16.079 26.079 P(72.5 X < 73.5)-Area in between 72.5 and 73.5 72.5-66 1.069171 1.07 73.5-66 1.233659 1.23 P(72.5X < 73.5) 11 = 66 ơ 6.079474 P(X < 73.5) 3.5 72.5 P(72.5 X < 73.5) P1.07Z<1.23) = P(Z < 1.23)-P(Z < 1.07) 0.8907- 0.8577 (from z-table)

- 0.0330 P(X=73) = 0.0330 Using excel instead of z-table gives: 0.0338312797 b) X ~ Binomial(n = 100, p = 0.56) Binomial can be approximated to normal with: μ-np-100 * 0.56-56 σ = V/np(1-р) _ V100 * (0.56) (1-0.56) 4.903869 4.964 P(52 X <59) Since we are approrimating a discrete binomial distribution by continuous nor- mal distribution, values between 51.5 and 52.5 both approzimate to 52. Thus between 52 and 59csponds to continuous noTmal distribution with P(51.5 X< 59.5) after continuity correction Normal Distribution, zı = 51.5,T2 = 59.5, μ = 56, σ = 4.964 We convert this to standard normal using z 51.5-56 ะ-0.906551 ะ-0.91 4.964 59.5-56 24.964 0.705095 0.71 P(51.5 X < 59.5) P(-0.91 < Z <0.71) = P(Z < 0.71)-P(Z <-0.91) 0.7611 0.1814 (from z-table) -0.5797 P(52 < X < 59) 0.5797 Using excel instead of z-table gives: 0.5773023269

XBinomial (n - 90,p- 0.1) ial can be approximated to normal with: np-90 * 0.41-36.9 μ p) = V/90 * (0.41) (1-0.41) 4.665940 4.666 P(X 2 40) Since we are approximating a discrete distribution by continuous normal distri bution, values between 39.5 and 40.5 both approrimate to 40. Thus, "greater than or equal to 40 " corresponds to continuous normal distribution with X > 39.5 after continuity coTrection. Normal Distribution, 39.5, μ-36.9, σ-4.666 We convert this to standard normal using z- 39.5 -36.9 4.666 0.557230.56 P(X > 39.5) = Area to the right of 39.5 u36.9 P(X> 39.5) - 4.66594 39.5 36.9 P(X > 39.5)-P(Z > 0.56)-1-P(Z < 0.56) = 1-0.7123 (from z-table) = 0.2877 P(X 2 40) 0.2877 Using excel instead of z-table gives: 0.2886852754

d) XBinomial (n - 105, p0.72) ial can be approximated to normal with: μ-np-105 * 0.72 = 75.6 σ = V np(1-p-V 105 * (0.72) (1-072) 4.600869 4.601 P(X 78) Since we are approximating a discrete distribution by continuous normal distri bution, values between 77.5 and 78.5 both approrimate to 78. Thus, "less than or equal to 78 ” coTTesponds to continuous normal distribution with X < 78.5 after continuity coTrection. Normal Distribution, x-78.5, μ-75.6, σ-4.601 We convert this to standard normal using z 78.5-75.6 4.601 0.630316 0.63 P(X < 78.5) - Area to the left of 78.5 μ-75.6 σ 4.600869 P(X78.5 78.5 75.6 P(X < 78.5)-P(Z < 0.63) = 0.7357 (from z-table) P(X < 78) = 0.7357 If excel is used instead of z-table, it would be: 0.7357559760