Question

# Week # Revenue Print Ad TV Ad 1 \$20,000 \$ 3,100 \$ 4,100 2 \$22,000 \$...

Week # Revenue Print Ad TV Ad

1 \$20,000 \$ 3,100 \$ 4,100

2 \$22,000 \$ 2,600 \$ 4,200

3 \$18,000 \$ 2,800 \$ 4,500

4 \$21,000 \$ 3,300 \$ 4,300

5 \$20,500 \$ 3,100 \$ 4,000

6 \$19,000 \$ 2,900 \$ 3,700

7 \$17,500 \$ 2,500 \$ 3,500

8 \$21,225 \$ 2,800 \$ 3,600

9 \$23,148 \$ 3,000 \$ 4,100

10 \$22,865 \$ 3,100 \$ 4,400

11 \$18,596 \$ 2,600 \$ 3,700

12 \$17,432 \$ 2,500 \$ 3,100

Using the given data as a basis, determine the probabilities that during the next 12 weeks revenue will exceed \$20,000, print ad will exceed \$3,000 and TV ad will exceed \$4,000.

using minitab

Descriptive Statistics: revenue, print add, tv add

Variable Mean StDev Variance Sum
revenue 20106 2012 4050094 241266
print add 2858.3 267.8 71742.4 34300.0
tv add 3933 416 173333 47200

P(x > 20000) = :

= P(z > -0.05268)

P(x > 20000) = 0.521 (using z table )

probabilities that during the next 12 weeks revenue will exceed \$20,000 is 0.521

P(x > 3000) =

=  0.2984

probabilities that during the next 12 weeks print ad will exceed \$3,000 is 0.2984

P(x > 4000) =

= = 0.436

probabilities that during the next 12 weeks  TV ad will exceed \$4,000 is 0.436

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