Normal Distribution, Solved Problems

Unit 6. Normal Distribution
Solution to problems

Statistics I. International Group Departamento de Economa Aplicada Universitat de Valncia

May 20, 2010

Problem 35

Random variable X : weekly ticket sales (units) of a museum. X ∼ N(1000, 180)

Find the probability of weekly sales exceeding 850 tickets. Find the probability of the interval 1000 to 1200 Take 5 weeks at random. Find the probability of weekly sales not exceeding 850 tickets in more than two weeks Ticket price is 4.5 Euros. Define the distribution of weekly revenue

Problem 35 (a)

Random variable X : weekly ticket sales (units) of a museum. X ∼ N(1000, 180) Find the probability of weekly sales exceeding 850 tickets. Find the probability of the interval 1000 to 1200 P(X > 850)? P(X > 850) = P(Z > 850 − 1000 ) = P(Z > −0.83) = 180

P(Z < 0.83) = P(Z < 0.83) = 0.7967

Problem 35 (a)
Random variable X : weekly ticket sales (units) of a museum. X ∼ N(1000, 180) Find the probability of weekly sales exceeding 850 tickets. Find the probability of the interval 1000 to 1200 P(1000 < X < 1200)? P(1000 < X < 1200) = P(X < 1200) − P(X < 1000) = 1200 − 1000 1000 − 1000 ) − P(Z < ) 180 180 = P(Z > −0.83) = P(Z < 1.11)−P(Z < 0) = 0.8665−0.5 = P(Z < = 0.3665

Problem 35 (b)
Probability of more than two weeks with sales not exceeding 850 tickets Y : number of weeks with sales not exceeding 850 tickets in 5 weeks. The probability of sales being less than 850 tickets (found in the first section) is 1-0.7967=0.2033. Let us analyze this random experiment Ticket sales for one week can be below or above 850 (success/failure) We repeat the previous step n = 5 trials Then the number of weeks with ticket sales below 850 in 5 trials, variable Y , follows a binomial distribution Y ∼ Bi(p = 0.2033, n = 5)

Problem 35 (b)

Probability of more than two weeks with sales not exceeding 850 tickets Y ∼ Bi(p = 0.2033, n = 5)

P(Y > 2)? P(Y > 2) = 1−P(Y ≤ 2) = 1−P(Y = 0)−P(Y = 1)−P(Y = 2) 5! 5! (0.2033)0