题目
Problem
As part of an investigation, Bobby collects a sample of 47 47 47 x x x a a a
(a) Find the range of these observations.
(1)
(b) Find the value of the median of these observations.
(1)
Given that the interquartile range of these observations is 0.31 0.31 0.31
(c) find the value of a a a
(3)
Bobby calculates the following statistics from these observations
∑ x = 23.72 , ∑ x 2 = 13.4228. \sum x=23.72,\qquad \sum x^2=13.4228. ∑ x = 23.72 , ∑ x 2 = 13.4228. (d) Show that the standard deviation of these observations is 0.176 0.176 0.176 3 3 3
(2)
Bobby now collects 18 18 18 y y y
(e) Using all 65 65 65 0.502 0.502 0.502 0.204 0.204 0.204
(i) Show that ∑ y = 8.91 \sum y=8.91 ∑ y = 8.91
(2)
(ii) Showing your working, calculate ∑ y 2 \sum y^2 ∑ y 2
(3)
题目中文翻译
Bobby 为一项调查收集了包含 47 47 47 x x x a a a 3 ∣ 2 3\mid2 3 ∣ 2 0.32 0.32 0.32
(a) 求这些观测值的极差。（1 分）
(b) 求这些观测值的中位数。（1 分）
已知这些观测值的四分位距是 0.31 0.31 0.31
(c) 求 a a a
Bobby 根据这些观测值算得 ∑ x = 23.72 \sum x=23.72 ∑ x = 23.72 ∑ x 2 = 13.4228 \sum x^2=13.4228 ∑ x 2 = 13.4228
(d) 证明这些观测值的标准差为 0.176 0.176 0.176
Bobby 又从同一项调查中收集了 18 18 18 y y y
(e) 使用全部 65 65 65 0.502 0.502 0.502 0.204 0.204 0.204
(i) 证明 ∑ y = 8.91 \sum y=8.91 ∑ y = 8.91
(ii) 写出计算过程，求 ∑ y 2 \sum y^2 ∑ y 2
解答
(a)
解法一
思路
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从 stem-and-leaf 读出最小值是 0.21 0.21 0.21 0.87 0.87 0.87
答题过程
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Range = 0.87 − 0.21 = 0.66. \begin{align*}
\text{Range}
=&\,0.87-0.21\\
=&\,0.66.
\end{align*} Range = = 0.87 − 0.21 0.66.
(b)
解法一
思路
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一共有 47 47 47 47 + 1 2 = 24 \dfrac{47+1}{2}=24 2 47 + 1 = 24 3 3 3 15 15 15 4 4 4 11 11 11 24 24 24 4 4 4 0.48 0.48 0.48
答题过程
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The median is the
47 + 1 2 = 24 th \begin{align*}
\frac{47+1}{2}=24\text{th}
\end{align*} 2 47 + 1 = 24 th observation.
The 24 24 24
0.48. \begin{align*}
0.48.
\end{align*} 0.48.
(c)
解法一
思路
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下四分位数位置是第 12 12 12 Q 1 = 0.35 Q_1=0.35 Q 1 = 0.35 I Q R = 0.31 IQR=0.31 I QR = 0.31
Q 3 = 0.35 + 0.31 = 0.66. \begin{align*}
Q_3=0.35+0.31=0.66.
\end{align*} Q 3 = 0.35 + 0.31 = 0.66. 上四分位数位置是第 36 36 36 0.60 + 0.01 a 0.60+0.01a 0.60 + 0.01 a a = 6 a=6 a = 6
答题过程
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The lower quartile is the 12 12 12
Q 1 = 0.35. \begin{align*}
Q_1=0.35.
\end{align*} Q 1 = 0.35. Since the interquartile range is 0.31 0.31 0.31
Q 3 − Q 1 = 0.31 Q 3 − 0.35 = 0.31 Q 3 = 0.66. \begin{align*}
Q_3-Q_1=&\,0.31\\
Q_3-0.35=&\,0.31\\
Q_3=&\,0.66.
\end{align*} Q 3 − Q 1 = Q 3 − 0.35 = Q 3 = 0.31 0.31 0.66. The upper quartile is the 36 36 36 0.60 + 0.01 a 0.60+0.01a 0.60 + 0.01 a
Therefore
0.60 + 0.01 a = 0.66 , \begin{align*}
0.60+0.01a=0.66,
\end{align*} 0.60 + 0.01 a = 0.66 , so
a = 6. \begin{align*}
a=6.
\end{align*} a = 6.
(d)
解法一
思路
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用标准差公式
s = ∑ x 2 n − ( ∑ x n ) 2 . \begin{align*}
s=\sqrt{\frac{\sum x^2}{n}-\Big(\frac{\sum x}{n}\Big)^2}.
\end{align*} s = n ∑ x 2 − ( n ∑ x ) 2 . 这里 n = 47 n=47 n = 47
答题过程
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First find the mean and mean square:
x ˉ = 23.72 47 = 0.504680851 … , ∑ x 2 47 = 13.4228 47 = 0.285591489 … . \begin{align*}
\bar{x}=&\,\frac{23.72}{47}=0.504680851\ldots,\\
\frac{\sum x^2}{47}=&\,\frac{13.4228}{47}=0.285591489\ldots.
\end{align*} x ˉ = 47 ∑ x 2 = 47 23.72 = 0.504680851 … , 47 13.4228 = 0.285591489 … . Hence
s = ∑ x 2 47 − x ˉ 2 = 0.030888727 … = 0.175751893 … . \begin{align*}
s=&\,\sqrt{\frac{\sum x^2}{47}-\bar{x}^2}\\
=&\,\sqrt{0.030888727\ldots}\\
=&\,0.175751893\ldots.
\end{align*} s = = = 47 ∑ x 2 − x ˉ 2 0.030888727 … 0.175751893 … . Therefore
s = 0.176 \begin{align*}
s=0.176
\end{align*} s = 0.176 to 3 3 3
解法二
思路
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先求离均差平方和 S x x S_{xx} S xx S x x / 47 \sqrt{S_{xx}/47} S xx /47
答题过程
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S x x = ∑ x 2 − ( ∑ x ) 2 47 = 13.4228 − 23.72 2 47 = 1.451770212 … . \begin{align*}
S_{xx}=&\,\sum x^2-\frac{(\sum x)^2}{47}\\
=&\,13.4228-\frac{23.72^2}{47}\\
=&\,1.451770212\ldots.
\end{align*} S xx = = = ∑ x 2 − 47 ( ∑ x ) 2 13.4228 − 47 23.7 2 2 1.451770212 … . Hence
s = S x x 47 = 1.451770212 … 47 = 0.175751893 … = 0.176 ( 3 s.f. ) . \begin{align*}
s=&\,\sqrt{\frac{S_{xx}}{47}}\\
=&\,\sqrt{\frac{1.451770212\ldots}{47}}\\
=&\,0.175751893\ldots\\
=&\,0.176\quad(3\text{ s.f.}).
\end{align*} s = = = = 47 S xx 47 1.451770212 … 0.175751893 … 0.176 ( 3 s.f. ) .
(e)(i)
解法一
思路
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全部 65 65 65 0.502 0.502 0.502 65 × 0.502 65\times0.502 65 × 0.502 47 47 47 x x x 18 18 18 y y y
答题过程
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For all 65 65 65
∑ x + ∑ y = 65 ( 0.502 ) . \begin{align*}
\sum x+\sum y=65(0.502).
\end{align*} ∑ x + ∑ y = 65 ( 0.502 ) . Using ∑ x = 23.72 \sum x=23.72 ∑ x = 23.72
∑ y = 65 ( 0.502 ) − 23.72 = 32.63 − 23.72 = 8.91. \begin{align*}
\sum y
=&\,65(0.502)-23.72\\
=&\,32.63-23.72\\
=&\,8.91.
\end{align*} ∑ y = = = 65 ( 0.502 ) − 23.72 32.63 − 23.72 8.91.
(e)(ii)
解法一
思路
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对全部 65 65 65
s 2 = ∑ x 2 + ∑ y 2 65 − x ˉ 2 . \begin{align*}
s^2=\frac{\sum x^2+\sum y^2}{65}-\bar{x}^2.
\end{align*} s 2 = 65 ∑ x 2 + ∑ y 2 − x ˉ 2 . 这里整体标准差是 0.204 0.204 0.204 0.502 0.502 0.502
答题过程
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For all 65 65 65
s 2 = ∑ x 2 + ∑ y 2 65 − x ˉ 2 . \begin{align*}
s^2=\frac{\sum x^2+\sum y^2}{65}-\bar{x}^2.
\end{align*} s 2 = 65 ∑ x 2 + ∑ y 2 − x ˉ 2 . Substitute the given values:
( 0.204 ) 2 = 13.4228 + ∑ y 2 65 − ( 0.502 ) 2 13.4228 + ∑ y 2 65 = ( 0.204 ) 2 + ( 0.502 ) 2 13.4228 + ∑ y 2 = 65 ( 0.041616 + 0.252004 ) ∑ y 2 = 65 ( 0.293620 ) − 13.4228 = 5.6625. \begin{align*}
(0.204)^2
=&\,\frac{13.4228+\sum y^2}{65}-(0.502)^2\\
\frac{13.4228+\sum y^2}{65}
=&\,(0.204)^2+(0.502)^2\\
13.4228+\sum y^2
=&\,65(0.041616+0.252004)\\
\sum y^2
=&\,65(0.293620)-13.4228\\
=&\,5.6625.
\end{align*} ( 0.204 ) 2 = 65 13.4228 + ∑ y 2 = 13.4228 + ∑ y 2 = ∑ y 2 = = 65 13.4228 + ∑ y 2 − ( 0.502 ) 2 ( 0.204 ) 2 + ( 0.502 ) 2 65 ( 0.041616 + 0.252004 ) 65 ( 0.293620 ) − 13.4228 5.6625. Therefore
∑ y 2 = 5.66 \begin{align*}
\sum y^2=5.66
\end{align*} ∑ y 2 = 5.66 to 3 3 3
