题目
Problem
A random sample of 100 carrots is taken from a farm and their lengths, L L L
Length, L L L Frequency, f f f Class mid point, x x x 5 ≤ L < 8 5\leq L<8 5 ≤ L < 8 5 6.5 8 ≤ L < 10 8\leq L<10 8 ≤ L < 10 13 9 10 ≤ L < 12 10\leq L<12 10 ≤ L < 12 16 11 12 ≤ L < 15 12\leq L<15 12 ≤ L < 15 25 13.5 15 ≤ L < 20 15\leq L<20 15 ≤ L < 20 30 17.5 20 ≤ L < 28 20\leq L<28 20 ≤ L < 28 11 24
A histogram is drawn to represent these data.
The bar representing the class 5 ≤ L < 8 5\leq L<8 5 ≤ L < 8
(a) Find the width and height of the bar representing the class 15 ≤ L < 20 15\leq L<20 15 ≤ L < 20
(3)
(b) Use linear interpolation to estimate the median length of these carrots.
(2)
(c) Estimate
(i) the mean length of these carrots,
(2)
(ii) the standard deviation of the lengths of these carrots.
(3)
A supermarket will only buy carrots with length between 9 cm and 22 cm.
(d) Estimate the proportion of carrots from the farm that the supermarket will buy.
(2)
Any carrots that the supermarket does not buy are sold as animal feed.
The farm makes a profit of 2.2 pence on each carrot sold to the supermarket, a profit of 0.8 pence on each carrot longer than 22 cm and a loss of 1.2 pence on each carrot shorter than 9 cm.
(e) Find an estimate of the mean profit per carrot made by the farm.
(2)
解答
(a)
解法一
思路
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直方图中面积代表频数。已知第一组的频数 5 对应图上面积 1.5 × 1 = 1.5 1.5\times1=1.5 1.5 × 1 = 1.5 30 ÷ 5 30\div5 30 ÷ 5
答题过程
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The class width 5 ≤ L < 8 5\leq L<8 5 ≤ L < 8
3 cm in data = 1.5 cm on the histogram . \begin{align*}
3\text{ cm in data}=1.5\text{ cm on the histogram}.
\end{align*} 3 cm in data = 1.5 cm on the histogram . For 15 ≤ L < 20 15\leq L<20 15 ≤ L < 20
5 × 1.5 3 = 2.5 cm . \begin{align*}
5\times\frac{1.5}{3}=2.5\text{ cm}.
\end{align*} 5 × 3 1.5 = 2.5 cm . The first bar has area
1.5 × 1 = 1.5 , \begin{align*}
1.5\times1=1.5,
\end{align*} 1.5 × 1 = 1.5 , and this represents frequency 5. The frequency 30 bar must therefore have area
1.5 × 30 5 = 9. \begin{align*}
1.5\times\frac{30}{5}=9.
\end{align*} 1.5 × 5 30 = 9. Hence its height is
9 2.5 = 3.6 cm . \begin{align*}
\frac{9}{2.5}=3.6\text{ cm}.
\end{align*} 2.5 9 = 3.6 cm .
(b)
解法一
思路
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总频数是 100，所以中位数在第 50 个数据附近。累计频数到 10 ≤ L < 12 10\leq L<12 10 ≤ L < 12 12 ≤ L < 15 12\leq L<15 12 ≤ L < 15
答题过程
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The cumulative frequency before the class 12 ≤ L < 15 12\leq L<15 12 ≤ L < 15
5 + 13 + 16 = 34. \begin{align*}
5+13+16=34.
\end{align*} 5 + 13 + 16 = 34. The median is the 50th value, so it is 50 − 34 = 16 50-34=16 50 − 34 = 16 12 ≤ L < 15 12\leq L<15 12 ≤ L < 15
Using linear interpolation,
median = 12 + 16 25 × 3 = 13.92. \text{median}
=12+\frac{16}{25}\times3
=13.92. median = 12 + 25 16 × 3 = 13.92. Therefore the median length is approximately
13.9 cm . \begin{align*}
13.9\text{ cm}.
\end{align*} 13.9 cm .
(c)(i)
解法一
思路
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分组数据估计平均数时，用组中值代表该组所有数据。
答题过程
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Using the class midpoints,
∑ f x = 5 ( 6.5 ) + 13 ( 9 ) + 16 ( 11 ) + 25 ( 13.5 ) + 30 ( 17.5 ) + 11 ( 24 ) = 1452. \begin{aligned}
\sum fx
=&\,5(6.5)+13(9)+16(11)\\
&\quad+25(13.5)+30(17.5)+11(24)\\
=&\,1452.
\end{aligned} ∑ f x = = 5 ( 6.5 ) + 13 ( 9 ) + 16 ( 11 ) + 25 ( 13.5 ) + 30 ( 17.5 ) + 11 ( 24 ) 1452. Hence
x ˉ = 1452 100 = 14.52. \begin{align*}
\bar x=\frac{1452}{100}=14.52.
\end{align*} x ˉ = 100 1452 = 14.52. So the estimated mean length is
14.5 cm . \begin{align*}
14.5\text{ cm}.
\end{align*} 14.5 cm .
(c)(ii)
解法一
思路
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继续使用组中值，先求 ∑ f x 2 \sum fx^2 ∑ f x 2
答题过程
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Using the class midpoints,
∑ f x 2 = 5 ( 6.5 2 ) + 13 ( 9 2 ) + 16 ( 11 2 ) + 25 ( 13.5 2 ) + 30 ( 17.5 2 ) + 11 ( 24 2 ) = 23280. \begin{aligned}
\sum fx^2
=&\,5(6.5^2)+13(9^2)+16(11^2)\\
&\quad+25(13.5^2)+30(17.5^2)+11(24^2)\\
=&\,23280.
\end{aligned} ∑ f x 2 = = 5 ( 6. 5 2 ) + 13 ( 9 2 ) + 16 ( 1 1 2 ) + 25 ( 13. 5 2 ) + 30 ( 17. 5 2 ) + 11 ( 2 4 2 ) 23280. Therefore
σ = ∑ f x 2 100 − x ˉ 2 = 23280 100 − 14.52 2 = 21.9696 = 4.687 … . \begin{aligned}
\sigma
=&\,\sqrt{\frac{\sum fx^2}{100}-\bar x^2}\\
=&\,\sqrt{\frac{23280}{100}-14.52^2}\\
=&\,\sqrt{21.9696}\\
=&\,4.687\ldots.
\end{aligned} σ = = = = 100 ∑ f x 2 − x ˉ 2 100 23280 − 14.5 2 2 21.9696 4.687 … . So the estimated standard deviation is
4.69 cm . \begin{align*}
4.69\text{ cm}.
\end{align*} 4.69 cm .
(d)
解法一
思路
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需要估计 9 ≤ L ≤ 22 9\leq L\leq22 9 ≤ L ≤ 22 8 ≤ L < 10 8\leq L<10 8 ≤ L < 10 20 ≤ L < 28 20\leq L<28 20 ≤ L < 28 20 20 20 22 22 22
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The estimated number of carrots with 9 ≤ L < 22 9\leq L<22 9 ≤ L < 22
1 2 ( 13 ) + 16 + 25 + 30 + 2 8 ( 11 ) . \begin{align*}
\frac{1}{2}(13)+16+25+30+\frac{2}{8}(11).
\end{align*} 2 1 ( 13 ) + 16 + 25 + 30 + 8 2 ( 11 ) . So
6.5 + 16 + 25 + 30 + 2.75 = 80.25. \begin{align*}
6.5+16+25+30+2.75=80.25.
\end{align*} 6.5 + 16 + 25 + 30 + 2.75 = 80.25. Therefore the estimated proportion is
80.25 100 = 0.8025. \begin{align*}
\frac{80.25}{100}=0.8025.
\end{align*} 100 80.25 = 0.8025.
(e)
解法一
思路
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按 (d) 的估计，买走的是 80.25 80.25 80.25 5 + 1 2 ( 13 ) = 11.5 5+\frac12(13)=11.5 5 + 2 1 ( 13 ) = 11.5 6 8 ( 11 ) = 8.25 \frac68(11)=8.25 8 6 ( 11 ) = 8.25
答题过程
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The estimated number of carrots sold to the supermarket is 80.25 80.25 80.25
The estimated number shorter than 9 cm is
5 + 1 2 ( 13 ) = 11.5. \begin{align*}
5+\frac{1}{2}(13)=11.5.
\end{align*} 5 + 2 1 ( 13 ) = 11.5. The estimated number longer than 22 cm is
6 8 ( 11 ) = 8.25. \begin{align*}
\frac{6}{8}(11)=8.25.
\end{align*} 8 6 ( 11 ) = 8.25. So the estimated mean profit per carrot is
2.2 ( 80.25 ) + 0.8 ( 8.25 ) − 1.2 ( 11.5 ) 100 = 176.55 + 6.6 − 13.8 100 = 1.6935. \begin{aligned}
\frac{2.2(80.25)+0.8(8.25)-1.2(11.5)}{100}
=&\,\frac{176.55+6.6-13.8}{100}\\
=&\,1.6935.
\end{aligned} 100 2.2 ( 80.25 ) + 0.8 ( 8.25 ) − 1.2 ( 11.5 ) = = 100 176.55 + 6.6 − 13.8 1.6935. Therefore the estimated mean profit is
1.7 pence per carrot . \begin{align*}
1.7\text{ pence per carrot}.
\end{align*} 1.7 pence per carrot . 