Question

Corn plants from a test plot are measured, and the distribution of heights at \(10-\mathrm{cm}\) intervals is recorded in the following table: $$\begin{array}{cc}\text { Height }(\mathrm{cm}) & \text { Plants (no.) } \\\100 & 20 \\\110 & 60 \\\120 & 90 \\\130 & 130 \\\140 & 180 \\\150 & 120 \\\160 & 70 \\\170 & 50 \\\180 & 40\end{array}$$ Calculate (a) the mean height, (b) the variance, (c) the standard deviation, and (d) the standard error of the mean. Plot a rough graph of plant height against frequency. Do the values represent a normal distribution? Based on your calculations, how would you assess the variation within this population?

Short answer

Short Answer: Plant heights in the given population have an approximate mean of 135.6 cm and a standard deviation of 23.47 cm. The standard error of the mean is 0.80 cm, suggesting an accurate estimation of the mean height for the entire population. By sketching a plot of plant height against frequency, we can observe if the distribution is approximately normal. Based on the calculated statistics and the shape of the distribution, we can conclude that there is a moderate amount of variation in height within this corn plant population.

Step-by-step solution

Calculate the Mean Height

To find the mean height, we need to find the sum of the product of each height and the number of plants at that height, then divide by the total number of plants. Using the given table, we can calculate the mean height \((\bar{x})\) using the formula: \(\bar{x} = \cfrac{\sum{(x \cdot f)}}{\sum{(f)}}\) $$ \bar{x} = \cfrac{(100\cdot20)+(110\cdot60)+(120\cdot90)+(130\cdot130)+(140\cdot180)+(150\cdot120)+(160\cdot70)+(170\cdot50)+(180\cdot40)}{20+60+90+130+180+120+70+50+40} $$ $$ \bar{x} = \cfrac{2000+6600+10800+16900+25200+18000+11200+8500+7200}{870} $$ $$ \bar{x} \approx 135.6 \ \text{cm} $$

Calculate the Variance

The variance \((s^2)\) can be calculated using the formula: \(s^2 = \cfrac{\sum{(x^2 \cdot f)}}{\sum{(f)}} - (\cfrac{\sum{(x \cdot f)}}{\sum{(f)}})^2\) $$ s^2 = \cfrac{(100^2 \cdot 20)+(110^2 \cdot 60)+(120^2 \cdot 90)+(130^2 \cdot 130)+(140^2 \cdot 180)+(150^2 \cdot 120)+(160^2 \cdot 70)+(170^2 \cdot 50)+(180^2 \cdot 40)}{870} - (135.6)^2 $$ $$ s^2 = \cfrac{200000+726000+1296000+2197000+3528000+2700000+1792000+1445000+1166400}{870} - 18362.56 $$ $$ s^2 \approx 550.9 $$

Calculate the Standard Deviation

The standard deviation \((s)\) is the square root of the variance: $$ s = \sqrt{s^2} = \sqrt{550.9} \approx 23.47 \ \text{cm} $$

Calculate the Standard Error of the Mean

Standard error of the mean (SEM) can be calculated using the formula: \(\text{SEM} = \cfrac{s}{\sqrt{N}}\) For our data, \(N = \sum{(f)} = 870\), so: $$ \text{SEM} = \cfrac{23.47}{\sqrt{870}} \approx 0.80 \ \text{cm} $$

Sketch the Plot of Plant Height Against Frequency

To create a rough sketch of the data, we can plot the given heights (x-axis) against the number of plants at each height (y-axis). The shape of this plot will help us determine if the data follows a normal distribution.

Determine Normality of the Distribution

By inspecting the plotted graph, we can observe if the data follows an approximate bell shape symmetric about the mean value. If the graph shows these characteristics, we can infer that the data distribution is approximately normal.

Assess the Variation Within the Population

In order to assess the variation within the population, we can examine the calculated statistics. The standard deviation (about \(23.47 \ \text{cm}\)) and the standard error of the mean (about \(0.80 \ \text{cm}\)) can provide insight on how much deviation there is from the mean and the accuracy of the mean estimation for the entire population. The standard deviation indicates a moderate amount of variability in plant heights; meanwhile, the standard error implies that the mean height is an accurate measure of the whole population. By carefully considering these statistics and the distribution's shape, an assessment of the variation within this corn plant population can be concluded.