Dylan Bautista -221/18 10: QMB 3200 Section UO7 Spring 2018 Time Remalning: 01:2

Question

Dylan Bautista -221/18 10: QMB 3200 Section UO7 Spring 2018 Time Remalning: 01:26:30 Sub Quiz: Quiz 2 (Take Home) This Question: 4 pts 5of13(°complete) This Quiz: 41 pts a. What is the probability that a randomly selected tax return retund will be more than $2200? (Round to four domal places as needed.) b. What is the probability hat a randomly selected tax return refund wil be between $1300 and $3000 (Round to four dearmal places as needed.) c. What is the probabilly that a randomly selected tax return refund will be between $3500 and $43007 D:Round to four decimal places as needed } d. What refund amount represents the 35th percentile of tax returns? S (Round to the nearest dolar as needed.) 0

Explanation / Answer

Solution:

Given in the question

Mean = $2754

Standard Deviation= $937

Solution(1):

P(X>2200) = 1- P(X<2200)

Z = (2200-2754)/937 = -0.59124

probability can be calculated from z table

P(X<2200) = 0.2776

P(X>2200) = 1-P(X<2200) = 1-0.2776 = 0.7224

Solution2:

P(1300<X<3000) = P(X<3000)-P(X<1300)

Z = (3000- 2754)/937 = 0.2625

P(X<3000) = 0.6026

Z= (1300-2754)/937 = -1.5517

P(X<1300) = 0.0606

P(1300<X<3000) = P(X<3000)-P(X<1300)

= 0.6026 - 0.0606 = 0.542 or 54.2%

Solution3:

P(3500<X<4300) = P(X<4300)-P(X<3500)

Z= (4300- 2754)/937 = 1.6499 or 1.65

P(X<4300) = 0.9505

Z = (3500-2754)/937 = 0.7961

P(X<3500) = 0.7852

P(3500<X<4300) = P(X<4300)-P(X<3500)

P(3500<X<4300) = 0.9505- 0.7852 = 0.1653 or 16.53%

Solution4:

given in the question p= 0.35

Z value can be calculated from z table i.e. -0.385

so tax refund value can be calculated as

-0.385 = (Xbar-2754)/937

-360.745 = xbar- 2754

Xbar = 2754-360.745 = 2393

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