In: Math
In a recent year the average movie ticket cost $10.50, In a random sample of 50 movie tickets from various areas
What is the probability that the mean cost exceeds $8.50, given that the population standard deviation is $1.50?
Solution :
mean = = 10.50
standard deviation = = 1.50
n = 50
= = 10.50
= / n = 1.50/ 50 = 0.21
P( > 8.50) = 1 - P( < 8.50)
= 1 - P[( - ) / < (8.50 - 10.50) / 0.21]
= 1 - P(z < -9.52)
Using z table,
= 1 - 0
= 1