Math Problem Statement

Find the​ range, variance, and standard deviation for the given sample​ data, if possible. If the measures of variation can be obtained for these​ values, do the results make​ sense? Biologists conducted experiments to determine whether a deficiency of carbon dioxide in the soil affects the phenotypes of peas. Listed below are the phenotype​ codes, where 1equals ​smooth-yellow, 2equals ​smooth-green, 3equals wrinkled ​yellow, and 4equals ​wrinkled-green. 1      1      2      2      3      2      4      1      2      2      2      2      1      2      2      3      1      2      2      2

Solution

Let's first organize the sample data and then calculate the range, variance, and standard deviation.

The sample data is:

1,1,2,2,3,2,4,1,2,2,2,2,1,2,2,3,1,2,2,21, 1, 2, 2, 3, 2, 4, 1, 2, 2, 2, 2, 1, 2, 2, 3, 1, 2, 2, 2

Step 1: Calculate the Range

The range is the difference between the maximum and minimum values in the data set.

  • Minimum value: 11
  • Maximum value: 44

Range=41=3\text{Range} = 4 - 1 = 3

Step 2: Calculate the Variance

Variance measures the spread of the data. The formula for sample variance is:

s2=1n1(xixˉ)2s^2 = \frac{1}{n-1} \sum (x_i - \bar{x})^2

Where:

  • nn is the sample size
  • xix_i are the individual data points
  • xˉ\bar{x} is the sample mean

Sample Size and Mean:

  • Sample size n=20n = 20
  • Mean xˉ=xin=1+1+2++220=4020=2\bar{x} = \frac{\sum x_i}{n} = \frac{1 + 1 + 2 + \cdots + 2}{20} = \frac{40}{20} = 2

Variance Calculation:

Now calculate the squared differences from the mean for each data point and sum them up:

(xixˉ)2=(12)2+(12)2+(22)2++(22)2\sum (x_i - \bar{x})^2 = (1-2)^2 + (1-2)^2 + (2-2)^2 + \cdots + (2-2)^2 =1+1+0+0+1+0+4+1+0+0+0+0+1+0+0+1+1+0+0+0=11= 1 + 1 + 0 + 0 + 1 + 0 + 4 + 1 + 0 + 0 + 0 + 0 + 1 + 0 + 0 + 1 + 1 + 0 + 0 + 0 = 11

Now divide by n1n - 1:

s2=11201=11190.579s^2 = \frac{11}{20 - 1} = \frac{11}{19} \approx 0.579

Step 3: Calculate the Standard Deviation

The standard deviation is the square root of the variance:

s=s2=0.5790.761s = \sqrt{s^2} = \sqrt{0.579} \approx 0.761

Conclusion

  • Range: 3
  • Variance: 0.579
  • Standard Deviation: 0.761

Does this make sense?

Yes, the values make sense. The range of 3 shows that the phenotype codes span a difference of three levels. The variance and standard deviation are relatively low, reflecting that the data points are clustered around the mean, indicating limited spread.

Do you want more details or have any questions?
Here are 5 relative questions:

  1. What is the interpretation of variance in a biological context?
  2. How would the data change if an additional phenotype code was introduced?
  3. What is the significance of the standard deviation in phenotypic experiments?
  4. Can range alone provide insights into data spread?
  5. How does sample size affect variance and standard deviation?

Tip: The standard deviation is always in the same units as the data, making it easier to interpret in context!

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Statistics
Measures of Variation
Range
Variance
Standard Deviation

Formulas

Range = max(data) - min(data)
Variance (sample) = (1 / (n - 1)) * Σ(x_i - x̄)²
Standard Deviation = √Variance

Theorems

Empirical Rule for Standard Deviation

Suitable Grade Level

Grades 10-12