题目
Problem
The lengths, x x x 50 50 50
Length Frequency 20 ⩽ x < 30 20\leqslant x<30 20 ⩽ x < 30 2 2 2 30 ⩽ x < 32 30\leqslant x<32 30 ⩽ x < 32 16 16 16 32 ⩽ x < 36 32\leqslant x<36 32 ⩽ x < 36 20 20 20 36 ⩽ x < 40 36\leqslant x<40 36 ⩽ x < 40 8 8 8 40 ⩽ x < 45 40\leqslant x<45 40 ⩽ x < 45 3 3 3 45 ⩽ x < 50 45\leqslant x<50 45 ⩽ x < 50 1 1 1
A histogram is drawn to represent these data.
The bar representing the class 32 ⩽ x < 36 32\leqslant x<36 32 ⩽ x < 36 2.5 2.5 2.5 7.5 7.5 7.5
(a) Calculate the width and the height of the bar representing the class 30 ⩽ x < 32 30\leqslant x<32 30 ⩽ x < 32
(3)
(b) Using linear interpolation, estimate the median of x x x
(2)
The weight, w w w 50 50 50
10 y = w − 20. 10y=w-20. 10 y = w − 20. These coded data are summarised by
∑ y = 104 , ∑ y 2 = 233.54. \sum y=104,\qquad \sum y^2=233.54. ∑ y = 104 , ∑ y 2 = 233.54. (c) Show that the mean of w w w 40.8 40.8 40.8
(2)
(d) Calculate the standard deviation of w w w
(4)
The weight of a pebble recorded as 40.8 40.8 40.8
(e) Without carrying out any further calculations, state, giving a reason, what effect this would have on the value of
(i) the mean of w w w
(ii) the standard deviation of w w w
(3)
解答
(a)
解法一
思路
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Histogram 里面积代表频数。已知 32 ⩽ x < 36 32\leqslant x<36 32 ⩽ x < 36 20 20 20 2.5 × 7.5 = 18.75 2.5\times7.5=18.75 2.5 × 7.5 = 18.75 30 ⩽ x < 32 30\leqslant x<32 30 ⩽ x < 32
答题过程
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For 32 ⩽ x < 36 32\leqslant x<36 32 ⩽ x < 36
2.5 ( 7.5 ) = 18.75. \begin{align*}
2.5(7.5)=18.75.
\end{align*} 2.5 ( 7.5 ) = 18.75. This represents a frequency of 20 20 20
For 30 ⩽ x < 32 30\leqslant x<32 30 ⩽ x < 32 16 16 16
16 20 ( 18.75 ) = 15. \begin{align*}
\frac{16}{20}(18.75)=15.
\end{align*} 20 16 ( 18.75 ) = 15. The class width 30 ⩽ x < 32 30\leqslant x<32 30 ⩽ x < 32 32 ⩽ x < 36 32\leqslant x<36 32 ⩽ x < 36
2 4 ( 2.5 ) = 1.25 cm . \begin{align*}
\frac{2}{4}(2.5)=1.25\text{ cm}.
\end{align*} 4 2 ( 2.5 ) = 1.25 cm . Therefore the height is
15 1.25 = 12 cm . \begin{align*}
\frac{15}{1.25}=12\text{ cm}.
\end{align*} 1.25 15 = 12 cm .
(b)
解法一
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50 50 50 25 25 25 32 32 32 18 18 18 32 ⩽ x < 36 32\leqslant x<36 32 ⩽ x < 36
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The median is the 25 25 25
The cumulative frequency before the class 32 ⩽ x < 36 32\leqslant x<36 32 ⩽ x < 36
2 + 16 = 18. \begin{align*}
2+16=18.
\end{align*} 2 + 16 = 18. Using linear interpolation,
median = 32 + 25 − 18 20 ( 36 − 32 ) = 32 + 7 20 ( 4 ) = 33.4. \begin{align*}
\text{median}
=&\,32+\frac{25-18}{20}(36-32)\\[3mm]
=&\,32+\frac{7}{20}(4)\\[3mm]
=&\,33.4.
\end{align*} median = = = 32 + 20 25 − 18 ( 36 − 32 ) 32 + 20 7 ( 4 ) 33.4. So the estimated median is
33.4 mm . \begin{align*}
33.4\text{ mm}.
\end{align*} 33.4 mm .
解法二
思路
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直观比例插值法。插值的本质是假设数据在区间内均匀分布。
中位数代表第 25 25 25 32 ⩽ x < 36 32\leqslant x<36 32 ⩽ x < 36 18 18 18 25 − 18 = 7 25 - 18 = 7 25 − 18 = 7 20 20 20 36 − 32 = 4 36 - 32 = 4 36 − 32 = 4
7 20 = 35 % \begin{align*}
\frac{7}{20} = 35\%
\end{align*} 20 7 = 35% 对应实际长度为：
35 % × 4 = 1.4 mm \begin{align*}
35\% \times 4 = 1.4\text{ mm}
\end{align*} 35% × 4 = 1.4 mm 所以中位数估算为：
32 + 1.4 = 33.4 mm \begin{align*}
32 + 1.4 = 33.4\text{ mm}
\end{align*} 32 + 1.4 = 33.4 mm 这种比例插值法避免了直接套用抽象公式，完全依靠数据在区间内均匀分布的直观占比来求解。
答题过程
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The median is the 25 25 25 [ 32 , 36 ) [32, 36) [ 32 , 36 ) 2 + 16 = 18 2 + 16 = 18 2 + 16 = 18 25 − 18 = 7 25 - 18 = 7 25 − 18 = 7 [ 32 , 36 ) [32, 36) [ 32 , 36 ) 20 20 20
The width of this interval is:
36 − 32 = 4 mm . \begin{align*}
36 - 32 = 4\text{ mm}.
\end{align*} 36 − 32 = 4 mm . The proportion of distance the median travels into this interval is:
7 20 = 0.35. \begin{align*}
\frac{7}{20} = 0.35.
\end{align*} 20 7 = 0.35. The distance into the interval is:
0.35 × 4 = 1.4 mm . \begin{align*}
0.35 \times 4 = 1.4\text{ mm}.
\end{align*} 0.35 × 4 = 1.4 mm . Thus, the estimated median is:
median = 32 + 1.4 = 33.4 mm . \begin{align*}
\text{median} =&\,\, 32 + 1.4\\[3mm]
=&\,\, 33.4\text{ mm}.
\end{align*} median = = 32 + 1.4 33.4 mm .
(c)
解法一
思路
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由 10 y = w − 20 10y=w-20 10 y = w − 20
w = 10 y + 20. \begin{align*}
w=10y+20.
\end{align*} w = 10 y + 20. 所以平均数也满足
w ˉ = 10 y ˉ + 20. \begin{align*}
\bar{w}=10\bar{y}+20.
\end{align*} w ˉ = 10 y ˉ + 20.
答题过程
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The mean of y y y
y ˉ = 104 50 = 2.08. \begin{align*}
\bar{y}=\frac{104}{50}=2.08.
\end{align*} y ˉ = 50 104 = 2.08. Since
w = 10 y + 20 , \begin{align*}
w=10y+20,
\end{align*} w = 10 y + 20 , we have
w ˉ = 10 y ˉ + 20 = 10 ( 2.08 ) + 20 = 40.8. \begin{align*}
\bar{w}
=&\,10\bar{y}+20\\[3mm]
=&\,10(2.08)+20\\[3mm]
=&\,40.8.
\end{align*} w ˉ = = = 10 y ˉ + 20 10 ( 2.08 ) + 20 40.8.
(d)
解法一
思路
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先求 y y y 10 10 10 w = 10 y + 20 w=10y+20 w = 10 y + 20 20 20 20 10 10 10 10 10 10
答题过程
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First,
Var ( Y ) = ∑ y 2 50 − ( ∑ y 50 ) 2 = 233.54 50 − ( 104 50 ) 2 = 4.6708 − 4.3264 = 0.3444. \begin{align*}
\operatorname{Var}(Y)
=&\,\frac{\sum y^2}{50}-\left(\frac{\sum y}{50}\right)^2\\[3mm]
=&\,\frac{233.54}{50}-\left(\frac{104}{50}\right)^2\\[3mm]
=&\,4.6708-4.3264\\[3mm]
=&\,0.3444.
\end{align*} Var ( Y ) = = = = 50 ∑ y 2 − ( 50 ∑ y ) 2 50 233.54 − ( 50 104 ) 2 4.6708 − 4.3264 0.3444. So
sd ( Y ) = 0.3444 . \begin{align*}
\operatorname{sd}(Y)=\sqrt{0.3444}.
\end{align*} sd ( Y ) = 0.3444 . Since w = 10 y + 20 w=10y+20 w = 10 y + 20
sd ( W ) = 10 sd ( Y ) = 10 0.3444 = 5.868 … . \begin{align*}
\operatorname{sd}(W)
=&\,10\operatorname{sd}(Y)\\[3mm]
=&\,10\sqrt{0.3444}\\[3mm]
=&\,5.868\ldots.
\end{align*} sd ( W ) = = = 10 sd ( Y ) 10 0.3444 5.868 … . Therefore
sd ( W ) = 5.87 g \begin{align*}
\operatorname{sd}(W)=5.87\text{ g}
\end{align*} sd ( W ) = 5.87 g to 3 3 3
解法二
思路
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原始统计量变换法。不先計算 y y y w = 10 y + 20 w = 10y + 20 w = 10 y + 20 w w w ∑ w \sum w ∑ w ∑ w 2 \sum w^2 ∑ w 2
∑ w = ∑ ( 10 y + 20 ) = 10 ∑ y + 20 n \begin{align*}
\sum w = \sum(10y + 20) = 10\sum y + 20n
\end{align*} ∑ w = ∑ ( 10 y + 20 ) = 10 ∑ y + 20 n 且
∑ w 2 = ∑ ( 10 y + 20 ) 2 = 100 ∑ y 2 + 400 ∑ y + 400 n \begin{align*}
\sum w^2 = \sum(10y + 20)^2 = 100\sum y^2 + 400\sum y + 400n
\end{align*} ∑ w 2 = ∑ ( 10 y + 20 ) 2 = 100 ∑ y 2 + 400 ∑ y + 400 n 分別代入已知的 ∑ y = 104 \sum y = 104 ∑ y = 104 ∑ y 2 = 233.54 \sum y^2 = 233.54 ∑ y 2 = 233.54 n = 50 n = 50 n = 50 sd ( W ) = ∑ w 2 n − w ˉ 2 \operatorname{sd}(W) = \sqrt{\frac{\sum w^2}{n} - \bar{w}^2} sd ( W ) = n ∑ w 2 − w ˉ 2
答题过程
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We transform the summary statistics from y y y w w w
Since w = 10 y + 20 w = 10y + 20 w = 10 y + 20 n = 50 n = 50 n = 50
∑ w = 10 ∑ y + 20 n = 10 ( 104 ) + 20 ( 50 ) = 1040 + 1000 = 2040 \begin{align*}
\sum w =&\,\, 10\sum y + 20n\\[3mm]
=&\,\, 10(104) + 20(50)\\[3mm]
=&\,\, 1040 + 1000\\[3mm]
=&\,\, 2040
\end{align*} ∑ w = = = = 10 ∑ y + 20 n 10 ( 104 ) + 20 ( 50 ) 1040 + 1000 2040 And for the sum of squares:
∑ w 2 = ∑ ( 10 y + 20 ) 2 = 100 ∑ y 2 + 400 ∑ y + 400 n = 100 ( 233.54 ) + 400 ( 104 ) + 400 ( 50 ) = 23354 + 41600 + 20000 = 84954 \begin{align*}
\sum w^2 =&\,\, \sum (10y + 20)^2\\[3mm]
=&\,\, 100\sum y^2 + 400\sum y + 400n\\[3mm]
=&\,\, 100(233.54) + 400(104) + 400(50)\\[3mm]
=&\,\, 23354 + 41600 + 20000\\[3mm]
=&\,\, 84954
\end{align*} ∑ w 2 = = = = = ∑ ( 10 y + 20 ) 2 100 ∑ y 2 + 400 ∑ y + 400 n 100 ( 233.54 ) + 400 ( 104 ) + 400 ( 50 ) 23354 + 41600 + 20000 84954 Now calculate the mean and variance of w w w
w ˉ = 2040 50 = 40.8 \begin{align*}
\bar{w} = \frac{2040}{50} = 40.8
\end{align*} w ˉ = 50 2040 = 40.8 Var ( W ) = ∑ w 2 n − w ˉ 2 = 84954 50 − ( 40.8 ) 2 = 1699.08 − 1664.64 = 34.44 \begin{align*}
\operatorname{Var}(W) =&\,\, \frac{\sum w^2}{n} - \bar{w}^2\\[3mm]
=&\,\, \frac{84954}{50} - (40.8)^2\\[3mm]
=&\,\, 1699.08 - 1664.64\\[3mm]
=&\,\, 34.44
\end{align*} Var ( W ) = = = = n ∑ w 2 − w ˉ 2 50 84954 − ( 40.8 ) 2 1699.08 − 1664.64 34.44 Finally, calculate the standard deviation:
sd ( W ) = 34.44 = 5.868 … \begin{align*}
\operatorname{sd}(W) =&\,\, \sqrt{34.44}\\[3mm]
=&\,\, 5.868\ldots
\end{align*} sd ( W ) = = 34.44 5.868 … So the standard deviation of w w w 5.87 5.87 5.87 3 3 3
(e)(i)
解法一
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新加入的数据正好等于原来的平均数 40.8 40.8 40.8
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The mean would not change, since the added value is equal to the current mean.
(e)(ii)
解法一
思路
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新加入的数据在正中心，不增加离均差；但数据总数增加了，所以整体离散程度会降低，标准差变小。
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The standard deviation would decrease, because the added value is at the mean, making the data more concentrated about the mean.
