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
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