Probability that the next student taking the test

Reference no: EM131069938

The Table of Random Numbers that appears in Appendix B of your text is arranged in blocks of five numbers each. Note that the arrangement in blocks is only for convenience, and has nothing to do with a relationship between the numbers.

Note: Students will be provided with a link to a replica of the table of random numbers so that the text is not required to be at hand.

1. Randomly pick three blocks of five numbers and write them down.

80181

16391

33238

48562

75767

40215

41274

56985

60688

82484

35908

19994

67715

19292

65411

2. What sampling method did you just use? Describe the rationale for your answer in a paragraph of text.

The Table of Random Numbers is arranged in blocks only of five numbers each for convenience only, there is no relationship between the numbers. A simple random sample was used to randomly pick three blocks of five numbers. In this type of sample each block of five numbers had an equal chance of being randomly selected.

3. What is the probability that the next student taking the test would happen to select the same three blocks of random numbers that you selected in number 1 above? Show your work so your instructor can provide feedback.

The population of the Table of Random Numbers consists of 100 blocks. The probability of the next student selecting the same block one time is 1/100 = 0.01 multiplied by three random selections (0.01 x 0.01 x 0.01) = 0.000001

The probability that the next student taking the test would happen to select the same three blocks of random numbers is 0.000001

4. Write down the last two digits of each of the 15 numbers selected in number 1 above. Present these numbers in an ordered stem and leaf plot.

Last two digits of 15 numbers selected: 81 91 38 62 67 15 74 85 88 84 08 94 15 92 11

Numbers placed in numerical order: 08 11 15 15 38 62 67 74 81 84 85 88 91 92 94

Ordered stem and leaf plot:

0 I 8

1 I 1 5 5

3 I 8

6 I 2 7

7 I 4

8 I 1 4 5 8

9 I 1 2 4

5. Using the data from number 4 above, construct a frequency distribution with an appropriate number of classes. Summarize the frequency distribution by presenting it in columns with headings. Include headings for Class, Frequency, Midpoint, Relative Frequency, and Cumulative Frequency. (4 points)

Class

0 - 19

20 - 39

40 - 59

60 - 79

80 - 99

Frequency (f)

Sum (∑f) = 15

Midpoint

9.5

29.5

49.5

69.5

89.5

Relative Freq.

0.27

0.07

0.2

0.47

∑f/n approx. 1.01

Cumulative Freq.

Mid-Point is the lower class limit plus the upper class limit divided by two.

0+19 = 19/2 = 9.5

20+39 = 59/2 = 29.5

40+59 = 99/2 = 49.5

60+79 = 139/2 = 69.5

80+99 = 179/2 = 89.5

Relative Frequency is the frequency divided by the sum of frequencies.

4/15 = approx. 0.27

1/15 = approx. 0.07

0/15 = 0

3/15 = 0.2

7/15 = approx. 0.47

Cumulative Frequency is the sum of frequencies of that class and all previous classes.

6. Describe the distribution created by the data in number 4 and 5 above in terms of its shape, modality, skewness, etc., using appropriate statistical terminology. (2 points)

The distribution created by the data can be described as an upside down graph curve that is skewed slightly towards the left end.

7. Calculate the mean, variance and standard deviation of the distribution of scores listed in number 4 above. Show your work so your instructor can provide feedback. (4 points)

Scores: 08 11 15 15 38 62 67 74 81 84 85 88 91 92 94 (Sum of scores = 905)

The mean of the scores is calculated by the sum of all scores divided by the total number of scores. 905/15 = approx. 60.34

The mean of the frequency distribution is calculated by sum of the midpoint of each class times the frequency of each class divided by the total number of scores. 902.5/15 = approx. 60.17

Midpoint

9.5

29.5

49.5

69.5

89.5

Frequency (f)

Sum (∑f) = 15

Midpoint times Frequency

29.5

208.5

626.5

Sum (∑x) = 902.5

The variance is calculated by ∑(x-µ)²/(n-1), where x represents each individual score.

(08-60.34)² = 2739.4756

(11-60.34)² = 2434.4356

(15-60.34)² = 2055.7156

(38-60.34)² = 499.0756

(62-60.34)² = 2.7556

(67-60.34)² = 44.3556

(74-60.34)² = 186.5956

(81-60.34)² = 426.8356

(84-60.34)² = 559.7956

(85-60.34)² = 608.1156

(88-60.34)² = 765.0756

(91-60.34)² = 940.0356

(92-60.34)² = 1002.3556

(94-60.34)² = 1132.9956

Sum = 15453.334 / 14 (n-1) = 1103.8096

The standard deviation is the square root of the variance.

Square root of 1103.8096 = 33.22

8. Calculate the 50th percentile of the distribution in number 4 above. What score is at the 50th percentile? Be sure to show your work. (3 points)

Numbers placed in numerical order: 08 11 15 15 38 62 67 74 81 84 85 88 91 92 94

The median is considered to be the 50^th percentile, a point in the data where 50% falls below and 50% falls above it. The 50^th percentile of the distribution set is 15 scores x 50% = 7.5 rounded up is 8. The score at the 50^th percentile is 74.

9. Based on what you know about the distribution above, what would be the probability of selecting a score at random that corresponds to a z-score of at least 1.4? (2 points)

A z-score of 1.4 is 0.9192 or 91.92%. The distribution above has two scores at least 91.92%, they are 92 and 92. 2 scores / 15 total score = 0.133 or approx. 13.3% probability of selecting a score at random that corresponds to a z-score of at least 1.4

Reference no: EM131069938

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