Question

# Let X represent a binomial random variable with n = 360 and p = 0.82. Find...

Let X represent a binomial random variable with n = 360 and p = 0.82. Find the following probabilities. (Do not round intermediate calculations. Round your final answers to 4 decimal places.

 Probability a. P(X ≤ 290) b. P(X > 300) c. P(295 ≤ X ≤ 305) d. P(X = 280)

Mean = np = 360 * 0.82 = 295.2

Standard deviation = sqrt (np(1-p))

= sqrt (360 * 0.82 ( 1 - 0.82) )

= 7.2894

a)

P(X <= 290) = P(Z < (290.5 - 295.2) / 7.2894) (With continuity correction)

= P(Z < -0.64)

= 0.2611 (From Z table)

b)

P(X > 300) = P(Z < (300.5 - 295.2) / 7.2894) (With continuity correction)

= P(Z < 0.73)

= 0.2327 (From Z table)

c)

P(295 <= X <= 305) = P(294.5 < X < 305.5)

= P(Z < (305.5 - 295.2) / 7.2894) - P(Z < (294.5 - 295.2) / 7.2894) (With continuity correction)

= P(Z < 1.41) - P(Z < -0.10)

= 0.9207 - 0.4602 (From Z table)

= 0.4605

d)

P(X = 280) = P(279.5 < X < 280.5)

= P(Z < (280.5 - 295.2) / 7.2894) - P(Z < (279.5 - 295.2) / 7.2894) (With continuity correction)

= P(Z < -2.02) - P(Z < -2.15)

= 0.0217 - 0.0158 (From Z table)

= 0.0059

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