In a recent year the average movie ticket cost $10.50, In a random sample of 50...

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?

Solutions

Expert Solution

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

milcah answered 2 months ago

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