Question #202058

A. The distribution of the height (X) in centimeter (CM) of the 16 teachers of SCNHS was presented below. construct a histogram for the random variable (X).

X | F

138 | 1

139 | 2

140 | 3

141 | 4

142 | 3

143 | 2

144 | 1

B. Use Empircal rule to complete the following table. write on the respective column the range or interval of the scores based on the given parameters.

        |MEAN | STANDARD DEVIATION | 68% | 95% | 99.7% |
      1 | 135 |    28              |     |     |       |
      2 | 87  |    5.5             |     |     |       |
      3 | 213 |    15              |     |     |       |
      4 | 567 |    20              |     |     |       |
      5 | 785 |    29              |     |     |       |
c. Illustrate the distribution in activity b through diagram.
1
Expert's answer
2021-06-04T13:39:10-0400

A.



B.

Empircal rule:

68% of the data falls within one standard deviation from the mean: μ±σ\mu\pm \sigma

95% of the data falls within two standard deviations from the mean : μ±2σ\mu\pm 2\sigma

99.7% of the data falls within one standard deviation from the mean: μ±3σ\mu\pm 3\sigma


1.

68%: 135±28=(107,163)135\pm 28=(107,163)

95%: 135±56=(79,191)135\pm 56=(79,191)

99.7%: 135±84=(51,219)135\pm 84=(51,219)


2.

68%: 87±5.5=(81.5,92.5)87\pm 5.5=(81.5,92.5)

95%: 87±11=(76,98)87\pm 11=(76,98)

99.7%: 87±16.5=(70.5,103.5)87\pm 16.5=(70.5,103.5)


3.

68%: 213±15=(198,228)213\pm 15=(198,228)

95%: 213±30=(183,243)213\pm 30=(183,243)

99.7%: 213±45=(168,258)213\pm 45=(168,258)


4.

68%: 567±20=(547,587)567\pm 20=(547,587)

95%: 567±40=(527,607)567\pm 40=(527,607)

99.7%: 567±60=(507,627)567\pm 60=(507,627)


5.

68%: 785±29=(756,814)785\pm 29=(756,814)

95%: 785±58=(727,843)785\pm 58=(727,843)

99.7%: 785±87=(698,872)785\pm 87=(698,872)








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