Question 1

1. a) The two variables under study are ‘Quality Rating’ and ‘Meal Price’.

‘Quality Rating’ is a qualitative variable. ‘Meal Price’ is a quantitative variable.

b)

 Meal Price (\$) Quality Rating [\$10 – \$15[ [\$15 – \$20[ [\$20 – \$24[ Totals Good 4 1 0 5 Very Good 1 4 1 6 Excellent 0 1 3 4 Totals 5 6 4 15

1. c) The chart that can be used to display meal prices is a histogram. This is suitable because the data we have is continuous data in which the prices can be divided into equal class widths and their relative frequency can be shown with the histogram.

d)

1. e)

Number of restaurants rated very good and above = 10

Total number of restaurants = 15

Percentage very good and above = (10/15) * 100

Percentage very good and above = 66.67%

f)

Number of restaurants with meals priced less than \$15 = 5

Total number of restaurants = 15

Percentage with meals priced less than \$15 = (5/15) * 100

Percentage with meals priced less than \$15 = 33.33%

Question 2

 X(1) 6 X(2) 7 X(3) 8 X(4) 9 X(5) 9 X(6) 9 X(7) 10 X(8) 13 X(9) 14 X(10) 15

1. a) Mean = sum of values / total number of values

Mean = (6 + 7 + 8 + 9 + 9 + 9 + 10 + 13 + 14 + 15) / 10

Mean = 100 / 10

Mean = 10

Since there is an even number of values, the median is the average of the two middle values which are X(5) and X(6).

Median = (9 + 9) / 2

Median = 9

The data set is positively skewed. This is because the mean is greater than the median value.

1. b) No there are no values which can be considered an outlier. Outliers are extreme data values which are not close to the other data values. In this data set we see that most of the values are close together.

1. c) The formula for sample standard deviation is:

n = 10

Mean = 10

x

X(1) 6 -4 16
X(2) 7 -3 9
X(3) 8 -2 4
X(4) 9 -1 1
X(5) 9 -1 1
X(6) 9 -1 1
X(7) 10 0 0
X(8) 13 3 9
X(9) 14 4 16
X(10) 15 5 25
TOTAL 0 82

s = 3.0185

1. d) The box plot of the given data shows that the left and right whiskers are almost equal and this indicates the possibility of symmetry. This data is positively skewed since the mean is greater than the median

Question 3

a)

The formula for expected value of X is denoted by:

 x p (x) x * p (x) 0 0.1 0 1 0.15 0.15 2 0.3 0.6 3 0.2 0.6 4 0.15 0.6 5 0.1 0.5 Total 1 2.45

Expected number of service calls = 2.45

1. b) The formula for standard deviation of a discrete random variable is:

 x p (x) x * p (x) x – u (x – u)^2 p(x)*(x – u)^2 0 0.1 0 -2.45 6.0025 0.60025 1 0.15 0.15 -1.45 2.1025 0.315375 2 0.3 0.6 -0.45 0.2025 0.06075 3 0.2 0.6 0.55 0.3025 0.0605 4 0.15 0.6 1.55 2.4025 0.360375 5 0.1 0.5 2.55 6.5025 0.65025 Total 1 2.45 0.3 17.515 2.0475

Standard Deviation = √2.0475 = 1.431

c)

P (X <= 0) = 0.1

P (X <= 1) = 0.1 + 0.15 = 0.25

P (X <= 2) = 0.1 + 0.15 + 0.3 = 0.55

P (X <= 3) = 0.1 + 0.15 + 0.3 + 0.2 = 0.75

P (X <= 4) = 0.1 + 0.15 + 0.3 + 0.2 + 0.15 = 0.90

P (X <= 5) = 0.1 + 0.15 + 0.3 + 0.2 + 0.15 + 0.1 = 1

1. d) We can see this value from the table:

P (X = 1) = 0.15

1. e) P (X >= 3) = P (X = 3) + P (X = 4) + P (X = 5)

P (X >= 3) = 0.2 + 0.15 + 0.1

P (X >= 3) = 0.45

Question 4

1. a)

Mean = 1550

Standard Deviation = 300

We use the following formula to find the z value for both values of x:

for x = 1000

z = (1000 – 1550) / 300

z = – 1.833

Then we see the normal distribution table and see that the corresponding probability value is 0.0336

for x = 2000

z = (2000 – 1550) / 300

z = 1.5

Then we see the normal distribution table and see that the corresponding probability value is 0.9332

Therefore,

P (1000 < X < 2000) = 0.9332 – 0.0336

P (1000 < X < 2000) = 0.8996

The probability is 0.8998 or 89.96%

1. b) We work backwards and see from the normal distribution table the value of z for which the probability is 0.95 since that is the upper 5%.

z = 1.645

Then we use the formula to get the value of x:

1.645 = (x – 1550) / 300

1.645 * 300 = x – 1550

x = 493.46 + 1550 = 2043.46

There would have to be 2044 fatal crashes for a year to be in the upper 5%.

Question 5

The formula to be used to calculate the confidence interval is:

n = 400

x = 152

p = 152 / 400 = 0.38

1 – p = 0.62

We see the value of z from the normal distribution table:

z = 1.96

Lower Limit = 0.33

Upper Limit = 0.43

The confidence interval is (0.33, 0.43)

The results show that approximately 33% to 43% of the job seekers change job because of higher compensation elsewhere. This is a high percentage and employers should try to address this issue.

Question 6

The formula to be used to calculate the confidence interval is:

n = 24

mean = x = 310.3

s = 23.8

We see the value of t from the t table:

t = 1.714

Lower Limit = 301.97

Upper Limit = 318.63

The confidence interval is (301.97, 318.63)

The results show that we can be 90% confident that the above mentioned confidence interval contains the true population mean. The average days worked of all payroll departments is within the values of 301.97 to 318.63

Question 7

The formula to be used to calculate the confidence interval is:

n1 = 90

n2 = 200

p1 = 72/90 = 0.8

p2 = 120/200 = 0.6

d = 0.8 – 0.6 = 0.2

We see the value of z from the normal distribution table:

z = 2.576

Lower Limit = 0.06

Upper Limit = 0.34

The confidence interval is (0.06, 0.34)

Question 8

The formula to be used to calculate the confidence interval is:

Then we use Excel interactive sheet to calculate the confidence interval.

 TO ESTIMATE A CONFIDENCE INTERVAL FOR A MEAN Confidence Level 95% INSTRUCTIONS SUMMARY STATISTICS Enter the Desired Confidence Level (e.g., 90%, 95%) in Cell D3 (Yellow). Then Enter Summary Statistics (n, Sample Mean, Sample Standard Deviation) In Cells D6 Through D8 (Yellow). Results Will Show in Green Cells. n 10 Sample Mean 70.000 Sample Standard Dev 17.500 RESULTS* t Using Table 3 2.262 Lower Limit 57.481 Upper Limit 82.519

The confidence interval is (57.48, 82.52)

The results show that we can be 95% confident that the above mentioned confidence interval contains the true population mean. The average professional football fan spends between 57.48 to 82.52 dollars on food at a single football game.

This can be solved manually as well:

n = 10

mean = x = 70

s = 17.5

We see the value of t from the t table:

t = 2.262

Lower Limit = 57.48

Upper Limit = 82.52

The confidence interval is (57.48, 82.52)

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