Business Data Analysis Questions Essay Example

BUSINESS DATA ANALYSIS7

Business Data Analysis

Question One

  1. D. reason 1.1 is not a valid probability.

Probabilities space range from 0 to 1

0 ≤ P (A) ≥ 1

  1. Mean of probability distribution is C.

Reason

The sum of the variable X and P(X)

From the distribution

= 0 x 0.2 + 1 x 0.4 + 2 x 0.3 + 3 x 0.1

= 0 + 0.4 + 0.6 + 0.3

= 1.3

  1. To get the variance

We square the variable X and multiply the result with P(x), we subtract the square of the mean from the result and then get a square root

= 02(0.2) + 12(0.4) + 22(0.3) + 32(0.1)

= 0 + 0.4 + 1.2 + 0.9

= 2.5

= 2.5 – 1.32

= 2.5 – 1.69

= 0.81

= 0.9 (A is the correct answer)

  1. P(Z>1.48)

1 — P (Z<= 1.48)

1 – 0.9306

0.0694

  1. P(X>5) = 0.4

  2. The probability of getting one four is

=1/6 x 1/6

= 1/36

  1. P(A) = 0.2 and P(B) = 0.3

The P (A and B) = P (A) P (B)

= 0.2 X 0.3

= 0.06 (C is the correct answer)

Question 2

  1. Using a z-score

= (890 – 945)/29

= -1.8966

= — 1.9

= 0.0287

  1. Getting two z-scores

= (960 – 945)/29

= 0.52

And

= (900 – 945)/29

= -1.55

Subtracting the 2 z-scores we get the needed probability

For 0.52 = 0.6985

And for -1.55 = 0.0606

The probability = 0.6985 – 0.0606

= 0.6379

  1. 5%, mean of 945 meters squared, standard deviation of 29 square meters

Probability = 0.05

= (5/100 x 945)/29

= 1.63

From z-table

= 0.9484

= 948 square meters

Question 3

  1. Binomial distribution

This is a binomial distribution because there are only 2 outcomes (the order is process online or not)

Let X = number of orders processed online

Here, n= 20, and x = 18. Let p= 0.95(processed online), q = 0.05 (not processed online)

The probability that 18 orders will be processed online:

P(X) = 0.95 x 18/20

= 0.855

  1. 2/20 x 5/100

= 0.1 x 0.05

= 0.005

  1. 185/200 x 0.95

= 0.879

Question 4

  1. Total number of patrons

= 308 male + 292 female

= 600

Total number of patrons who order dessert

= 90 male + 46 Female

= 136

Probability that a patron orders dessert

= 136/600

= 17/75

  1. Probability of a female patron ordering dessert

= 46/600

= 23/300

17/75 x 23/300

= 182/600

= 91/300

  1. 17/75

  2. The events are independent

This is because if a male patron orders a dessert, it does not exclude a female patron from ordering the dessert

Question 5

  1. Rate of bankruptcy 5.5 per week

For 5 businesses to go bankrupt

P(A>5)

= (1/5.5) x 5

= 0.9

  1. 20 businesses in 4 weeks

Going bankrupt

P (A=20)

= (1/5.5) x 20/4

= 0.91