**Chapter 7. Distributions**

**1.You are watching: What is the probability that z is between 0.0 and 2.0** Prepare a table showing the frequencies, percentages, and also cumulative percentages for the following data. Make the bins 5 units large and make the first bin have actually the lower limit the 41.

89, 69, 76, 88, 74, 41, 49, 68, 72, 83, 66, 69, 70, 77,

56, 61, 58, 56, 63, 61, 66, 74, 69, 76, 68, 72, 83, 56

Value | Frequency | Percent | Cum. Percent |

41-45 | 1 | 3.57% | 3.57% |

46-50 | 1 | 3.57% | 7.14% |

51-55 | 0 | 0.00% | 7.14% |

56-60 | 4 | 14.29% | 21.43% |

61-65 | 3 | 10.71% | 32.14% |

66-70 | 8 | 28.57% | 60.71% |

71-75 | 4 | 14.29% | 75.00% |

76-80 | 3 | 10.71% | 85.71% |

81-85 | 2 | 7.14% | 92.86% |

86-90 | 2 | 7.14% | 100.00% |

Total | 28 | 100.00% |

**2.** attract a histogram because that the outcomes of #1. Make the bins 5 units wide and do the an initial bin have the reduced limit the 41.

**3.** draw histograms for the following varieties of distributions:

a. Symmetric single-peaked

b. Symmetric bimodal

c. Asymmetric single-peaked

d. Asymmetric bimodal distribution

e. Rectangular (flat) distribution

**4.** listed below are the final exam scores in percentages because that students in a statistics course that supplies a various text.

male | |

81.61 | 56.66 |

83.68 | 75.30 |

59.51 | 84.02 |

70.79 | 60.25 |

63.45 | 89.66 |

64.99 | 77.15 |

62.88 | 86.19 |

84.59 | 66.80 |

82.94 | 88.48 |

38.82 | 73.54 |

75.39 | 64.06 |

69.35 | 51.72 |

66.76 | 93.92 |

65.19 | 87.65 |

73.22 | 90.34 |

69.50 | 89.62 |

89.24 | 44.35 |

74.92 | 72.00 |

83.21 | 54.10 |

83.23 | 90.46 |

82.79 | 77.64 |

89.03 | 62.54 |

72.34 | 82.96 |

**a. Construct different frequency tables because that males and also females. Set them up so they have 12 bins.**

Females | |||

Value | Frequency | Percent | Cum. Percent |

35.00-39.99 | 0 | 0.00% | 0.00% |

40.00-44.99 | 1 | 4.55% | 4.55% |

45.00-49.99 | 0 | 0.00% | 4.55% |

50.00-54.99 | 1 | 4.55% | 9.09% |

55.00-59.99 | 0 | 0.00% | 9.09% |

60.00-64.99 | 1 | 4.55% | 13.64% |

65.00-69.99 | 3 | 13.64% | 27.27% |

70.00-74.99 | 4 | 18.18% | 45.45% |

75.00-79.99 | 1 | 4.55% | 50.00% |

80.00-84.99 | 4 | 18.18% | 68.18% |

85.00-89.99 | 4 | 18.18% | 86.36% |

90.00-94.99 | 3 | 13.64% | 100.00% |

Total | 22 | 100.00% |

Males | |||

Value | Frequency | Percent | Cum. Percent |

35.00-39.99 | 1 | 4.17% | 4.17% |

40.00-44.99 | 0 | 0.00% | 4.17% |

45.00-49.99 | 0 | 0.00% | 4.17% |

50.00-54.99 | 1 | 4.17% | 8.33% |

55.00-59.99 | 2 | 8.33% | 16.67% |

60.00-64.99 | 5 | 20.83% | 37.50% |

65.00-69.99 | 2 | 8.33% | 45.83% |

70.00-74.99 | 2 | 8.33% | 54.17% |

75.00-79.99 | 3 | 12.50% | 66.67% |

80.00-84.99 | 5 | 20.83% | 87.50% |

85.00-89.99 | 3 | 12.50% | 100.00% |

90.00-94.99 | 0 | 0.00% | 100.00% |

Total | 24 | 100.00% |

b. Draw a histogram that shows both distributions, choose the one ~ above the appropriate side of page 58.

c. Execute women do much better than men? What deserve to you say around this by assessing the table and also histogram girlfriend made?

d. Attract the cumulative portion curves because that the two teams on the histograms, making use of red because that females and also blue because that males.

e. Find the medians because that the two teams by assessing the cumulative percentage curves. What carry out the medians tell girlfriend in solution to question 4 c?

The typical for females is 75.0 and also the average for males is about 67.0. The females it seems ~ to have done much better than the males.

f. Space either of this distributions normal? How deserve to you tell? If lock aren\"t, what is it about them that renders them no normal?

Neither is normal. The circulation for males is bimodal and the one because that females is asymmetric and skewed.

**5.** assume the mean is 72 and also the traditional deviation is 2.4.

a. What percent of cases lie in between the mean and 2.0 standard deviations over the mean?

This snapshot shows the area between the mean (z = 0.0) and also z = 2.0. You have the right to see in the table the this area is .4772 of the total area under the common curve. In a normal distribution, 47.72% that all cases have z-scores in between 0.0 and 2.0.

z | z come mean | smaller area | larger area |

0.00 | 0.0000 | 0.5000 | 0.5000 |

: | : | : | : |

1.90 | 0.4713 | 0.0287 | 0.9713 |

2.00 | 0.4772 | 0.0228 | 0.9772 |

b. What percent of cases are an ext than 1.8 standard deviations away from the mean?

Do the right fifty percent of the distribution first. You desire the ones over z =1.8, therefore you use the table to acquire the \"smaller area\" because that z = 1.8, i m sorry is .0359. You likewise want the area listed below 1.8, which is the exact same as the area above z = 1.8, therefore you main point .0359 by 2 and find that 7.18% are more than 1.8 standard deviations away from the mean.

z | z come mean | smaller area | larger area |

0.00 | 0.0000 | 0.5000 | 0.5000 |

: | : | : | : |

1.80 | 0.4641 | 0.0359 | 0.9641 |

1.90 | 0.4713 | 0.0287 | 0.9713 |

2.00 | 0.4772 | 0.0228 | 0.9772 |

c. What percent of instances are an ext than 1.2 and less 보다 1.8 conventional deviations below the mean?

First convert the ranges to z-scores. This is easy:

-1.2 and also -1.8.

Then look for both of this on the table to obtain the areas. Native the median to z = 1.2 is .3849; indigenous the average to z = 1.8 is .4641. Subtract the very first from the second:

.4641 - .3849 = .0792

to view that 7.92% of cases are between 1.2 and also 1.8 conventional deviations below the mean.

z | z to mean | smaller area | larger area |

0.00 | 0.0000 | 0.5000 | 0.5000 |

: | : | : | : |

1.00 | 0.3413 | 0.1587 | 0.8413 |

1.10 | 0.3643 | 0.1357 | 0.8643 |

1.20 | 0.3849 | 0.1151 | 0.8849 |

: | : | : | : |

1.80 | 0.4641 | 0.0359 | 0.9641 |

1.90 | 0.4713 | 0.0287 | 0.9713 |

2.00 | 0.4772 | 0.0228 | 0.9772 |

d. If you desire the peak fifteen percent the scores, exactly how far above the mean do you have to go?

This time you begin with the area (15%) and need the z-score that synchronizes to it. The z-score will certainly divide the regular curve into two sections; the smaller will be the optimal 15% and also the larger will be the other 85%. For this reason look for the number in the \"smaller area\" obelisk closest to 0.15 and also you view 1.04 is the best. You have to be at least 1.04 typical deviations above the median to be in the optimal 15%.

z | z to mean | smaller area | larger area |

: | : | : | : |

1.00 | 0.3413 | 0.1587 | 0.8413 |

1.01 | 0.3438 | 0.1562 | 0.8438 |

1.02 | 0.3461 | 0.1539 | 0.8461 |

1.03 | 0.3485 | 0.1515 | 0.8485 |

1.04 | 0.3508 | 0.1492 | 0.8508 |

1.05 | 0.3531 | 0.1469 | 0.8531 |

**6.** If the scores top top an exam are usually distributed, the median exam score is 86.4, the typical deviation is 11.5, and the sample dimension is 500, exactly how many human being have scores over 95?

First, calculation the z-score. It would be:

95 - 86.4 = 8.6

and then:

8.6 /11.5 = .7478

The \"smaller area\" for z = .75 is 0.2266 or 22.66%. Finally, 22.66% of 500 is 113.3 which rounds to 113 people.

z | z to mean | smaller area | larger area |

: | : | : | : |

0.73 | 0.2673 | 0.2327 | 0.7673 |

0.74 | 0.2704 | 0.2296 | 0.7704 |

0.75 | 0.2734 | 0.2266 | 0.7734 |

0.76 | 0.2764 | 0.2236 | 0.7764 |

0.77 | 0.2794 | 0.2206 | 0.7794 |

7. If the scores on an exam are usually distributed, the median is 83.1, standard deviation is 13, and sample dimension is 600, just how many people have scores between 95 and 105?

First, calculation the z-scores. For 95, that is:

(95 - 83.1) / 13 = .9154 = z

For 105, it is:

(105 - 83.1) / 13 = 1.685 = z

For life scores that 95 and also 105, the z-scores will certainly be .9154 and 1.685. The area because that z = .92 is .3212; the area for z = 1.69 is .4545, for this reason to find the area between 95 and 105, subtract 32.12% native 45.45% to obtain 13.33%. To discover out how many people, multiply 13.33% by 600 to acquire 79.98, which rounds to 80 people.

z | z come mean | smaller area | larger area |

: | : | : | : |

0.90 | 0.3159 | 0.1841 | 0.8159 |

0.91 | 0.3186 | 0.1814 | 0.8186 |

0.92 | 0.3212 | 0.1788 | 0.8212 |

0.93 | 0.3238 | 0.1762 | 0.8238 |

: | : | : | : |

1.67 | 0.4525 | 0.0475 | 0.9525 |

1.68 | 0.4535 | 0.0465 | 0.9535 |

1.69 | 0.4545 | 0.0455 | 0.9545 |

1.70 | 0.4554 | 0.0446 | 0.9554 |

**8. What is the z-score the the human being in the 80th percentile?**

**This time you begin with the area (80%) and need the z-score that corresponds to it. The z-score will certainly divide the regular curve right into two sections; the larger will it is in the bottom 80% and also the smaller will be the other 20%. So look because that the number in the \"larger area\" obelisk closest come 0.80 and you view 0.84 is the best. You need to be at the very least .84 typical deviations above the mean to it is in in the 80th percentile or the optimal 20%.See more: What'S It Worth? How Much Is A Confederate 20 Dollar Bill Worth**

z | z to mean | smaller area | larger area |

: | : | : | : |

0.83 | 0.2967 | 0.2033 | 0.7967 |

0.84 | 0.2995 | 0.2005 | 0.7995 |

0.85 | 0.3023 | 0.1977 | 0.8023 |

0.86 | 0.3051 | 0.1949 | 0.8051 |

0.87 | 0.3078 | 0.1922 | 0.8078 |

**9. Does the answer to question 8 count on the mean and standard deviation? Why or why not?**

**10. If her z-score is 1.85, what percentile room you in?**

**Since the bigger area for 1.85 is .9678, 96.78% of everyone else is below you. This method you are in the 97th percentile.**

z | z to mean | smaller area | larger area |

: | : | : | : |

1.84 | 0.4671 | 0.0329 | 0.9671 |

1.85 | 0.4678 | 0.0322 | 0.9678 |

1.86 | 0.4686 | 0.0314 | 0.9686 |