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QuestionGiven heights of 200 fir trees, create a cumulative frequency table, curve, and estimate median, IQR, mean, SD, and variance.

Studdy Solution

STEP 1

Assumptions1. The heights of the fir trees are given in the table. . The heights are grouped into intervals.
3. The frequency represents the number of trees in each height interval.

STEP 2

To construct a cumulative frequency table, we start by copying the original table and adding a new row for the cumulative frequency.
\begin{tabular}{|l|c|c|c|c|c|c|c|} \hline Height, h(m)h(m) & 0<h≤10<h \leq1 & 1<h≤21<h \leq2 & 2<h≤2<h \leq & <h≤4<h \leq4 & 4<h≤54<h \leq5 & 5<h≤105<h \leq10 & Cumulative Frequency \\ \hline Frequency &17 &35 &69 &51 &22 &5 & \\ \hline\end{tabular}

STEP 3

The cumulative frequency for each height interval is the sum of the frequencies of that interval and all previous intervals. For the first interval, the cumulative frequency is the same as the frequency.
Cumulative Frequency0<h≤1=Frequency0<h≤1=17Cumulative\, Frequency_{0<h \leq1} = Frequency_{0<h \leq1} =17

STEP 4

For the second interval, add the frequency of the second interval to the cumulative frequency of the first interval.
Cumulative Frequency1<h≤2=Cumulative Frequency0<h≤1+Frequency1<h≤2=17+35Cumulative\, Frequency_{1<h \leq2} = Cumulative\, Frequency_{0<h \leq1} + Frequency_{1<h \leq2} =17 +35

STEP 5

Calculate the cumulative frequency for the second interval.
Cumulative Frequency1<h≤2=17+35=52Cumulative\, Frequency_{1<h \leq2} =17 +35 =52

STEP 6

Repeat this process for the remaining intervals. The cumulative frequency for the third interval isCumulative Frequency2<h≤3=Cumulative Frequency1<h≤2+Frequency2<h≤3=52+69Cumulative\, Frequency_{2<h \leq3} = Cumulative\, Frequency_{1<h \leq2} + Frequency_{2<h \leq3} =52 +69

STEP 7

Calculate the cumulative frequency for the third interval.
Cumulative Frequency2<h≤3=52+69=121Cumulative\, Frequency_{2<h \leq3} =52 +69 =121

STEP 8

Continue this process for the remaining intervals. The cumulative frequencies for the fourth, fifth, and sixth intervals areCumulative Frequency3<h≤4=Cumulative Frequency2<h≤3+Frequency3<h≤4=121+51Cumulative\, Frequency_{3<h \leq4} = Cumulative\, Frequency_{2<h \leq3} + Frequency_{3<h \leq4} =121 +51Cumulative Frequency4<h≤5=Cumulative Frequency3<h≤4+Frequency4<h≤5=172+22Cumulative\, Frequency_{4<h \leq5} = Cumulative\, Frequency_{3<h \leq4} + Frequency_{4<h \leq5} =172 +22Cumulative Frequency5<h≤10=Cumulative Frequency4<h≤5+Frequency5<h≤10=194+5Cumulative\, Frequency_{5<h \leq10} = Cumulative\, Frequency_{4<h \leq5} + Frequency_{5<h \leq10} =194 +5

STEP 9

Calculate the cumulative frequencies for the fourth, fifth, and sixth intervals.
Cumulative Frequency3<h≤4=121+51=172Cumulative\, Frequency_{3<h \leq4} =121 +51 =172Cumulative Frequency4<h≤5=172+22=194Cumulative\, Frequency_{4<h \leq5} =172 +22 =194Cumulative Frequency5<h≤=194+5=199Cumulative\, Frequency_{5<h \leq} =194 +5 =199

STEP 10

Now, the cumulative frequency table is complete.
\begin{tabular}{|l|c|c|c|c|c|c|c|} \hline Height, h(m)h(m) & 0<h≤0<h \leq & <h≤2<h \leq2 & 2<h≤32<h \leq3 & 3<h≤43<h \leq4 & 4<h≤54<h \leq5 & 5<h≤105<h \leq10 & Cumulative Frequency \\ \hline Frequency &17 &35 &69 &51 &22 &5 & \\ \hline Cumulative Frequency &17 &52 &121 &172 &194 &199 & \\ \hline\end{tabular}

STEP 11

To draw a cumulative frequency curve, plot the cumulative frequencies on the y-axis against the upper limit of each height interval on the x-axis. Connect the points with a smooth curve.

STEP 12

To estimate the median height of the fir trees, find the height that corresponds to the cumulative frequency that is halfway between the minimum and maximum cumulative frequencies. In this case, the median height corresponds to a cumulative frequency of100 (since the total number of trees is200).

STEP 13

Looking at the cumulative frequency table, we see that a cumulative frequency of100 falls in the interval 2<h≤32<h \leq3. Therefore, the median height is estimated to be in this interval.

STEP 14

The interquartile range is the range between the first quartile (Q) and the third quartile (Q3). To find these quartiles, we need to find the heights that correspond to cumulative frequencies of50 (Q) and150 (Q3).

STEP 15

Looking at the cumulative frequency table, we see that a cumulative frequency of50 falls in the interval <h≤2<h \leq2 and a cumulative frequency of150 falls in the interval 3<h≤43<h \leq4. Therefore, the interquartile range is estimated to be between these two intervals.

STEP 16

The10th percentile is the height below which10% of the observations fall. To find this, we need to find the height that corresponds to a cumulative frequency of20 (since10% of200 is20).

STEP 17

Looking at the cumulative frequency table, we see that a cumulative frequency of20 falls in the interval 0<h≤0<h \leq. Therefore, the10th percentile height is estimated to be in this interval.

STEP 18

To calculate the mean, we need to multiply each height by its frequency, sum these products, and then divide by the total number of trees. However, since the heights are given as intervals, we will use the midpoint of each interval as the representative height.

STEP 19

The midpoints of the height intervals are.5,1.5,.5,3.5,4.5, and7.5. The mean is then calculated asMean=∑i=16(Midpointi×Frequencyi)Total Number of TreesMean = \frac{\sum_{i=1}^{6} (Midpoint_i \times Frequency_i)}{Total\, Number\, of\, Trees}

STEP 20

Substitute the midpoints, frequencies, and total number of trees into the formula to calculate the mean.
Mean=(0.5×17)+(.5×35)+(.5×69)+(3.5×51)+(4.5×22)+(7.5×5)200Mean = \frac{(0.5 \times17) + (.5 \times35) + (.5 \times69) + (3.5 \times51) + (4.5 \times22) + (7.5 \times5)}{200}

STEP 21

Calculate the mean.
Mean=(8.5)+(52.5)+(172.5)+(178.5)+(99)+(37.5)200=548.5200=.7425Mean = \frac{(8.5) + (52.5) + (172.5) + (178.5) + (99) + (37.5)}{200} = \frac{548.5}{200} =.7425

STEP 22

The standard deviation is a measure of the dispersion of the heights. It is calculated as the square root of the variance. The variance is calculated asVariance=∑i=16((Midpointi−Mean)×Frequencyi)Total Number of TreesVariance = \frac{\sum_{i=1}^{6} ((Midpoint_i - Mean)^ \times Frequency_i)}{Total\, Number\, of\, Trees}

STEP 23

Substitute the midpoints, mean, frequencies, and total number of trees into the formula to calculate the variance.
Variance=((0.5−.7425)×17)+((1.5−.7425)×35)+((.5−.7425)×69)+((3.5−.7425)×51)+((.5−.7425)×22)+((7.5−.7425)×5)200Variance = \frac{((0.5 -.7425)^ \times17) + ((1.5 -.7425)^ \times35) + ((.5 -.7425)^ \times69) + ((3.5 -.7425)^ \times51) + ((.5 -.7425)^ \times22) + ((7.5 -.7425)^ \times5)}{200}

STEP 24

Calculate the variance.
Variance=(.06×17)+(1.56×35)+(0.06×69)+(0.56×51)+(3.06×22)+(22.56×)200=1.189Variance = \frac{(.06 \times17) + (1.56 \times35) + (0.06 \times69) + (0.56 \times51) + (3.06 \times22) + (22.56 \times)}{200} =1.189

STEP 25

The standard deviation is the square root of the variance.
Standard Deviation=Variance=1.189Standard\, Deviation = \sqrt{Variance} = \sqrt{1.189}

STEP 26

Calculate the standard deviation.
Standard Deviation=1.189=1.09Standard\, Deviation = \sqrt{1.189} =1.09

STEP 27

To draw a box and whisker plot, we need the minimum, first quartile (Q1), median, third quartile (Q3), and maximum values. The minimum and maximum values are the lower and upper limits of the height intervals, which are0 and10. The first quartile, median, and third quartile were estimated in steps13 and15.

STEP 28

Draw a number line from0 to10. Draw a box from Q1 to Q3 and draw a line through the box at the median. Draw whiskers from the box to the minimum and maximum values.

STEP 29

The box and whisker plot represents the distribution of the heights of the fir trees.

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