The Following Data Were Obtained From A Survey Of College Students Th
The following data were obtained from a survey of college students. The variable X represents the number of non-assigned books read during the past six months. The probabilities associated with different values of X are provided as follows:
P(X = 0) = 0.55
P(X = 1) = 0.15
P(X = 2) = 0.10
P(X = 3) = 0.10
P(X = 4) = 0.04
P(X = 5) = 0.03
P(X = 6) = 0.03
Calculate the variance of the variable X, rounded to two decimal places.
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The task requires calculating the variance of a discrete random variable X, which represents the number of non-assigned books read during the past six months. Variance is a measure of dispersion indicating how much the values of the variable are spread out from the mean. The calculation involves two primary steps: determining the expected value (mean) of the variable and then computing the expected value of the squared deviations from this mean.
Given the probability distribution for X, the first step is to compute the mean (µ) of the distribution. The mean is calculated as the sum of each value of X multiplied by its corresponding probability:
µ = Σ [x * P(X = x)]
Using the provided data:
µ = (0)(0.55) + (1)(0.15) + (2)(0.10) + (3)(0.10) + (4)(0.04) + (5)(0.03) + (6)(0.03)
Calculating each term:
0 * 0.55 = 0.0
1 * 0.15 = 0.15
2 * 0.10 = 0.20
3 * 0.10 = 0.30
4 * 0.04 = 0.16
5 * 0.03 = 0.15
6 * 0.03 = 0.18
Adding these, the mean µ is:
µ = 0.0 + 0.15 + 0.20 + 0.30 + 0.16 + 0.15 + 0.18 = 1.34
Next, the variance (σ²) is computed using the formula:
σ² = E[(X - µ)²] = Σ [ (x - µ)² * P(X = x) ]
We calculate each squared deviation for each value of X and multiply by its probability:
For x=0: (0 - 1.34)² * 0.55 = (1.7956) * 0.55 ≈ 0.9876
For x=1: (1 - 1.34)² * 0.15 = (0.1156) * 0.15 ≈ 0.0173
For x=2: (2 - 1.34)² * 0.10 = (0.4356) * 0.10 ≈ 0.0436
For x=3: (3 - 1.34)² * 0.10 = (2.7556) * 0.10 ≈ 0.2756
For x=4: (4 - 1.34)² * 0.04 = (7.0556) * 0.04 ≈ 0.2822
For x=5: (5 - 1.34)² * 0.03 = (13.2356) * 0.03 ≈ 0.3971
For x=6: (6 - 1.34)² * 0.03 = (21.4156) * 0.03 ≈ 0.6425
Adding up these individual components yields the variance:
Variance ≈ 0.9876 + 0.0173 + 0.0436 + 0.2756 + 0.2822 + 0.3971 + 0.6425 = 2.6489
Rounding to two decimal places, the variance of X is approximately 2.65.
In conclusion, the variance quantifies the dispersion of the number of non-assigned books read among students over six months. A variance of 2.65 indicates that while most students read a number close to the average (approximately 1.34), there is some variability in their reading behaviors. This statistical measure
offers valuable insights into reading habits and can inform targeted interventions to promote reading among students.
References
Allen, M. (2017).
Probability, Statistics, and Data Analysis for Engineering and the Sciences . Cengage Learning.
Devore, J. L. (2015).
Probability and Statistics for Engineering and the Sciences . Cengage Learning.
Moore, D. S., Notz, W. I., & Fligner, M. A. (2013).
Statistics: The Environment and Data
. W. H. Freeman.
Rice, J. A. (2006).
Mathematical Statistics and Data Analysis
. Cengage Learning.
Wasserman, L. (2004).
All of Statistics: A Concise Course in Statistical Inference . Springer.
Snedecor, G. W., & Cochran, W. G. (1989).
Statistical Methods
. Iowa State University Press.
Mendenhall, W., Beaver, R. J., & Beaver, B. M. (2012).
Introduction to Probability and Statistics
. Brooks/Cole.
Freund, J. E. (2010).
Modern Elementary Statistics
. Pearson.
Lehmann, E. L., & Casella, G. (2003).
Theory of Point Estimation . Springer.
Rubin, D. B. (2004).
Multiple Imputation for Nonresponse in Surveys . Wiley.