The owner of a computer repair shop has determined that their daily revenue has mean $7200 and standard deviation $1200. The daily revenue totals for the next 30 days will be monitored. What is the probability that the mean daily revenue for the next 30 days will exceed $7000? Round to four decimal places.

0.0853

Step-by-step explanation:

Step 1

The first step would be to find the Z score

Formula for Z score is given as

z = x – μ /( σ ÷ √n)

Where

x = observed value

μ = mean

σ = Standard deviation

n = sample size

In the question we were given the above values as :

x = observed value = $7500

μ = mean = $7200

σ = Standard deviation = $1200

n = samples size = 30

Z score (z) = $7500 – $7200 /($1200÷ √30)

Z score = $300/ ( $1200 ÷ √30)

Z score = 1.3693063938

Approximately, the Z score = 1.37

Step 2

Using this formula:

1 - P(z > 1.37)

Using the Standard distribution table,

P(z > 1.37) is given as

0.9146565

Therefore,

1 - P(z > 1.37)

= 1 - 0.9146565

= 0.0853435

In the question we were asked to round it up to 4 decimal places

= 0.0853

Therefore, the probability that the mean daily revenue for the next 30 days will exceed $7000 is 0.0853.

the correct answer to your question is a. 4^4

good luck my friend!

hope this !

maybe if i you could me by giving me brainliest?

! < 3

have a splendid day < 3 : )