# Normal distribution case

Problem 9 (page 168-169)

Mean (μ) = 0.8 seconds

Your Time

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Standard deviation (σ) = 0.2 seconds

Sample size (S) = 30 (random sample)

What is the probability that the sample mean is less than 0.75 seconds? - **Normal distribution case** introduction.?

Z = (value – mean) / (standard deviation)

= (0.75 – 0.8)/0.2

= – 0.25

Approximate cumulative area for the standard normal distribution for the z value of -0.25 is 0.4015, which is the probability (40.15%) that the sample mean is less than 0.75 seconds.

What is the probability that the sample mean is between 0.70 and 0.90 second?

Z value at 0.70

Z = (0.70 – 0.8)/0.2

= – 0.5

Cumulative area for z value – 0.5 is 0.309.

Z value at 0.90

Z = (0.90 – 0.8)/0.2

= 0.5

Cumulative area for z value 0.5 is 0.692.

Therefore, area between 0.70 and 0.90 second is = 0.692 – 0.309 = 0.383

The probability that the sample mean is between 0.70 and 0.90 second will be 0.383 (38.3%).

The probability is 80% that the sample mean is between what two values symmetrically distributed around the population mean?

Given probability = 0.8 (symmetrically distributed around the population mean)

Therefore the area will be difference between 90% area and 10% area.

Z value at cumulative area of 0.1 is – 2.3333 and Z value at cumulative area of 0.9 is 1.2833.

– 2.3333 = (value1 – 0.8)/0.2

à value1 = 0.33334 seconds

1.2833 = (value1 – 0.8)/0.2

à value1 = 1.05666 seconds

Therefore, the probability will be 80% when the sample mean is between 0.33 and 1.06 seconds (approximately) symmetrically distributed around the population mean

The probability is 90% that the sample mean is less than what value?

Z value at cumulative area of 0.9 is 1.2833

1.2833 = (value – 0.8)/0.2

à value1 = 1.05666 seconds

The probability is 90% when the sample mean is less than 1.06 seconds (approximately).

Problem 17 (page 149-150)

Time

Location(office)

Time

Location(office)

1.48

1

7.55

2

1.75

1

3.75

2

0.78

1

0.1

2

2.85

1

1.1

2

0.52

1

0.6

2

1.6

1

0.52

2

4.15

1

3.3

2

3.97

1

2.1

2

1.48

1

0.58

2

3.1

1

4.02

2

1.02

1

3.75

2

0.53

1

0.65

2

0.93

1

1.92

2

1.6

1

0.6

2

0.8

1

1.53

2

1.05

1

4.23

2

6.32

1

0.08

2

3.93

1

1.48

2

5.45

1

1.65

2

0.97

1

0.72

2

Mean

2.214

Mean

2.0115

Median

1.540

Median

1.505

Standard deviation

1.718038661

Standard deviation

1.891705649

Combined Office Location 1 and 2

Mean

2.113

Standard deviation

1.787

Median

1.505

Office Location 1

n

20

Mean

2.214

Median

1.540

95% CI

1.410

to 3.018

95.9% CI

0.970

to 3.100

SE

0.3842

Range

5.80

Variance

2.952

IQR

2.638

SD

1.718

95% CI

1.307

to 2.509

Percentile

0th

0.520

(minimum)

CV

77.6%

25th

0.947

(1st quartile)

50th

1.540

(median)

Skewness

1.13

75th

3.584

(3rd quartile)

Kurtosis

0.29

100th

6.320

(maximum)

Shapiro-Wilk W

0.84

p

0.004

From the histogram for office location 1, it is visible that the data appear to be normally distributed. However, there is positive skewness of 1.13. From the Normality Plot, it is visible that data are approximately normally distributed.

Office Location 2

n

20

Mean

2.012

Median

1.505

95% CI

1.126

to 2.897

95.9% CI

0.600

to 3.300

SE

0.4230

Range

7.47

Variance

3.579

IQR

2.963

SD

1.892

95% CI

1.439

to 2.763

Percentile

0th

0.080

(minimum)

CV

94.0%

25th

0.600

(1st quartile)

50th

1.505

(median)

Skewness

1.47

75th

3.563

(3rd quartile)

Kurtosis

2.41

100th

7.550

(maximum)

Shapiro-Wilk W

0.84

p

0.004

From the histogram for office location 2, it is visible that the data appear to be normally distributed. However, there is positive skewness of 1.47. From the Normality Plot, it is visible that data are approximately normally distributed.

Office Location 1 and 2 (combined)

n

40

Mean

2.113

Median

1.505

95% CI

1.541

to 2.684

96.2% CI

0.970

to 2.100

SE

0.2825

Range

7.47

Variance

3.192

IQR

2.818

SD

1.787

95% CI

1.464

to 2.294

Percentile

0th

0.080

(minimum)

CV

84.6%

25th

0.745

(1st quartile)

50th

1.505

(median)

Skewness

1.25

75th

3.563

(3rd quartile)

Kurtosis

1.14

100th

7.550

(maximum)

Shapiro-Wilk W

0.86

p

0.000

From the histogram for office location 1 and 2 combined, it is visible that the data appear to be normally distributed. However, there is positive skewness of 1.25. Which is less because this time the sample size is big (40) as compared to earlier ones (20). From the Normality Plot, it is visible that data are approximately normally distributed.

Positive Skewness: A distribution curve that is not symmetrical is referred to as skewed or asymmetrical (distribution) curve and skewness is a measure of a distribution from symmetry. A probability distribution is said to be positively skewed if the mass of the distribution in concentrated on the left of the figure or if the largest portion of data is below the mean. In this case, the right tail will be the longest and the distribution is said to be right skewed.