Question

The owner of a computer repair shop has determined that their daily revenue has mean $7200 and standard deviation $1200. The daily revenue is normally distributed. a) What is the probability that a randomly selected day will have a revenue of at most $7000? b) The daily revenue for the next 30 days will be monitored. What is the probability that the mean daily revenue for the next 30 days will exceed $7500?

Answer #1

Solution :

Given that ,

mean = = $7200

standard deviation = =$1200

a)

P(x $7000)

= P[(x - ) / (7000 - 7200) /1200 ]

= P(z -0.17)

= 0.4325

probability = 0.4325

b)

P(x > $7500) = 1 - P(x < 7500)

= 1 - P((x - ) / < (7500 - 7200) / 1200)

= 1 - P(z < 0.25)

= 1 - 0.5987

= 0.4013

Probability = 0.4013

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