偏度练习题 - 掌握偏度的判断和计算
以下是4道综合练习题，涵盖偏度的判断、计算和分析。
In a survey of the earnings of some college students who worked weekend jobs, the median wage was $36.50. The 75th percentile was $45.75 and the interquartile range was $30.50. Use the quartiles to describe the skewness of the distributions.
解答过程：
A group of estate agents recorded the time spent on the first meeting with a random sample of 120 of their clients. The mean time spent with their clients is 31.1 minutes and the variance is 78.05. The median time is 29.7 minutes and \( Q_1 \) and \( Q_3 \) values are 25.8 minutes and 34.8 minutes.
One measure of skewness is found using \( \frac{3(\text{mean} - \text{median})}{\text{standard deviation}} \)
a) Evaluate this measure and describe the skewness of the data
b) State, giving a reason, which you would recommend for them to use.
解答过程：
a) 偏度计算：
b) 推荐统计量：
The following stem and leaf diagram summarises the wing length, to the nearest mm, of a random sample of 67 birds.
| Wing length | Leaf | Key: \( 5 \mid 0 = 50 \text{mm} \) |
|---|---|---|
| 5 | 0 0 0 1 1 2 2 3 3 3 4 4 | (12) |
| 5 | 5 5 6 6 6 7 8 8 9 9 | (10) |
| 6 | 0 1 1 1 3 3 4 4 4 4 | (10) |
| 6 | 5 5 6 7 8 9 9 | (7) |
| 7 | 1 1 2 2 3 3 | (6) |
| 7 | 5 7 9 9 | (4) |
| 8 | 1 1 1 2 2 3 3 4 | (8) |
| 8 | 7 8 9 | (3) |
| 9 | 0 1 1 2 | (4) |
| 9 | 5 7 9 | (3) |
a) Write down the mode.
b) Find the median and quartiles of the data.
c) Construct a box plot to represent the data.
d) Comment on the skewness of the distribution.
e) Calculate the mean and standard deviation for the data.
f) Use another method to show that the data is skewed.
g) State, giving a reason, which of b or e would you recommend using to summarise the data in the diagram.
解答过程：
a) 众数：茎6的叶4出现4次，故众数是64mm
b) 四分位数：总数据量67
d) 偏度评论：
g) 推荐统计量：
An orange farmer randomly selects 120 oranges from her farm. The histogram below shows the diameters (in mm) of the oranges.
Calculate an estimate of the mean and standard deviation. Comment on why the mean is only an estimate, whether there any outliers, and the type of skewness displayed by the histogram.
| Diameter (mm) | Frequency | Midpoint |
|---|---|---|
| 60-70 | 10 | 65 |
| 70-80 | 35 | 75 |
| 80-90 | 55 | 85 |
| 90-100 | 20 | 95 |
解答过程：
均值计算：
\[ \text{均值} = \frac{65×10 + 75×35 + 85×55 + 95×20}{120} ≈ 82.08 \, \text{mm} \]
标准差计算：
\[ \Sigma fx^2 = 65^2×10 + 75^2×35 + 85^2×55 + 95^2×20 = 824750 \]
\[ \text{方差} = \frac{824750}{120} - (82.08)^2 ≈ 135.79, \quad \text{标准差} ≈ 11.65 \, \text{mm} \]
评论：