Question

In: Statistics and Probability

Table 6.0 shows a sample of the maximum capacity (maximum number of spectators) of sports stadiums....

Table 6.0 shows a sample of the maximum capacity (maximum number of spectators) of sports stadiums. The table does not include horse-racing or motor-racing stadiums.

40,000

40,000

45,050

45,500

46,249

48,134

49,133

50,071

50,096

50,466

50,832

51,500

51,500

51,900

52,000

52,132

52,200

52,530

52,692

53,864

54,000

55,000

55,000

55,000

55,000

55,000

55,000

55,082

57,000

58,008

59,680

60,000

60,000

60,492

60,580

62,380

62,872

64,035

65,000

65,050

65,647

66,000

66,161

67,428

68,349

68,976

69,372

70,107

70,585

71,594

72,000

72,922

73,379

74,500

75,025

76,212

78,000

80,000

80,000

82,300

Table 6.0

  1. Calculate the sample mean and the sample standard deviation for the maximum capacity of sports stadiums (the data).

- sd = 60136 and 10468

b. Construct a histogram.

c. Draw a smooth curve through the midpoints of the tops of the bars of the histogram.

d. In words, describe the shape of your histogram and smooth curve.

e. Let the sample mean approximate μ and the sample standard deviation approximate σ. The distribution of X can then be approximated by X ~ _____(_____,_____).

f. Use the distribution in part e to calculate the probability that the maximum capacity of sports stadiums is less than 67,000 spectators.

g. Determine the cumulative relative frequency that the maximum capacity of sports stadiums is less than 67,000 spectators. Hint: Order the data and count the sports stadiums that have a maximum capacity less than 67,000. Divide by the total number of sports stadiums in the sample.

h. Why aren’t the answers to part f and part g exactly the same?

Solutions

Expert Solution

We find the Mean and standard deviation of the maximum seating capacity of stadiums using Minitab

a. Mean = 60143

Standard Deviation = 10462

b.

c. The smooth curve over histogram

d. The histogram is bell shaped, however slightly skewed to the right.

e. X can be more or less approximated by a normal distribution.

X ~ N(60143,10462)

f. P[ X < 67000]

= P[(X - 60143)/10462 < (67000 - 60143)/10462]

= P[ Z < 0.6554] {Z = (X - 60143)/10462 ~ N(0,1)}

= 0.74389 {Values obtained from a standard normal table}

g.

To determine the cumulative relative frequency that maximum capacity of sports stadiums is less than 67000 spectators.

= (Number of stadiums in sample with stadium capacity less than 67000)/ (Total number of stadiums in sample)

= 43/60

= 0.7167

h. The answers in f and g aren't exactly the same because in f we assume normality and fit a normal distribution to the data obtained and then calculate the probability. However in g we just calculate the exact probability based on sample values that are given in the data.


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