37. Standard deviation problems

37. Standard deviation problems - step by step solution procedure and detailed explanation for each problem.

Below are 10 standard deviation problems, each with a step-by-step solution procedure and a detailed explanation. The problems range from basic to more complex, covering various scenarios to illustrate the concept of standard deviation, which measures the dispersion of data points around the mean. Each problem includes the problem statement, solution steps, and an explanation of the reasoning, formulas, and their applications.

Problem 1: Standard Deviation of a Small Dataset

Problem: Find the standard deviation of the dataset: 2, 4, 6, 8, 10.

Solution Procedure:

Calculate the mean:
- Sum of data:
  2 + 4 + 6 + 8 + 10 = 30
  .
- Number of data points (( n )): 5.
- Mean (
  \mu
  ):
  \frac{30}{5} = 6
  .
Calculate the squared deviations from the mean:
- For each data point
  x_i
  , compute
  (x_i - \mu)^2
  :
  - (2 - 6)^2 = (-4)^2 = 16
  - (4 - 6)^2 = (-2)^2 = 4
  - (6 - 6)^2 = 0^2 = 0
  - (8 - 6)^2 = 2^2 = 4
  - (10 - 6)^2 = 4^2 = 16
Sum the squared deviations:
- 16 + 4 + 0 + 4 + 16 = 40
  .
Calculate the variance (population variance, since no sample is specified):
- Variance (
  \sigma^2
  ):
  \frac{\sum (x_i - \mu)^2}{n} = \frac{40}{5} = 8
  .
Calculate the standard deviation:
- Standard deviation (
  \sigma
  ):
  \sqrt{\sigma^2} = \sqrt{8} \approx 2.828
  .
State the answer: The standard deviation is approximately 2.83 (rounded to two decimal places).

Detailed Explanation:

Why standard deviation? It measures how spread out the data points are from the mean. A larger standard deviation indicates greater variability.
Population vs. sample: Since the problem does not specify a sample, we assume the dataset represents the entire population, using
\sigma^2 = \frac{\sum (x_i - \mu)^2}{n}
.
Formula breakdown: The variance is the average of squared deviations, and the standard deviation is its square root, returning to the original units.
Verification: The values are evenly spaced (step of 2), and the standard deviation reflects moderate spread around the mean of 6.

Answer:

\sigma \approx 2.83

Problem 2: Sample Standard Deviation

Problem: A sample of test scores is 85, 90, 92, 88, 95. Find the sample standard deviation.

Solution Procedure:

Calculate the mean:
- Sum:
  85 + 90 + 92 + 88 + 95 = 450
  .
- Number of data points (( n )): 5.
- Mean (
  \bar{x}
  ):
  \frac{450}{5} = 90
  .
Calculate the squared deviations:
- (85 - 90)^2 = (-5)^2 = 25
- (90 - 90)^2 = 0^2 = 0
- (92 - 90)^2 = 2^2 = 4
- (88 - 90)^2 = (-2)^2 = 4
- (95 - 90)^2 = 5^2 = 25
Sum the squared deviations:
- 25 + 0 + 4 + 4 + 25 = 58
  .
Calculate the sample variance:
- Sample variance (
  s^2
  ):
  \frac{\sum (x_i - \bar{x})^2}{n-1} = \frac{58}{5-1} = \frac{58}{4} = 14.5
  .
Calculate the sample standard deviation:
- s = \sqrt{14.5} \approx 3.807
  .
State the answer: The sample standard deviation is approximately 3.81.

Detailed Explanation:

Sample vs. population: The problem specifies a sample, so we use
s^2 = \frac{\sum (x_i - \bar{x})^2}{n-1}
, dividing by
n-1
to correct for bias in estimating the population variance.
Why
n-1
? This is Bessel’s correction, accounting for the fact that the sample mean is used instead of the true population mean, reducing underestimation.
Interpretation: A standard deviation of 3.81 indicates the scores vary moderately around the mean of 90.

Answer:

s \approx 3.81

Problem 3: Standard Deviation Using the Computational Formula

Problem: Find the standard deviation of the dataset: 3, 7, 8, 12 using the computational formula.

Solution Procedure:

Use the computational formula:
\sigma = \sqrt{\frac{\sum x_i^2}{n} - \left(\frac{\sum x_i}{n}\right)^2}
.
Calculate sums:
- \sum x_i = 3 + 7 + 8 + 12 = 30
  .
- \sum x_i^2 = 3^2 + 7^2 + 8^2 + 12^2 = 9 + 49 + 64 + 144 = 266
  .
- n = 4
  .
Compute the mean:
- \mu = \frac{30}{4} = 7.5
  .
Apply the formula:
- \frac{\sum x_i^2}{n} = \frac{266}{4} = 66.5
  .
- \left(\frac{\sum x_i}{n}\right)^2 = (7.5)^2 = 56.25
  .
- Variance:
  \sigma^2 = 66.5 - 56.25 = 10.25
  .
- Standard deviation:
  \sigma = \sqrt{10.25} \approx 3.202
  .
State the answer: The standard deviation is approximately 3.20.

Detailed Explanation:

Computational formula: This is equivalent to the definitional formula but avoids calculating deviations, making it computationally efficient:
\sigma^2 = \frac{\sum x_i^2}{n} - \mu^2
.
Why use it? It’s useful for large datasets or when deviations are cumbersome to compute manually.
Verification: Using the definitional formula (as in Problem 1) yields the same variance:
\sum (x_i - 7.5)^2 = 41
,
\sigma^2 = \frac{41}{4} = 10.25
.

Answer:

\sigma \approx 3.20

Problem 4: Standard Deviation of a Frequency Distribution

Problem: A dataset is given as a frequency table:

Value ( ( x ) )	Frequency ( ( f ) )
10	2
20	3
30	5
Find the standard deviation.

Solution Procedure:

Calculate the mean:
- Total frequency (( n )):
  2 + 3 + 5 = 10
  .
- Sum of
  x \cdot f
  :
  (10 \cdot 2) + (20 \cdot 3) + (30 \cdot 5) = 20 + 60 + 150 = 230
  .
- Mean (
  \mu
  ):
  \frac{230}{10} = 23
  .
Calculate the squared deviations weighted by frequency:
- For each value:
  f \cdot (x - \mu)^2
  :
  - x = 10
    :
    (10 - 23)^2 = (-13)^2 = 169
    ,
    2 \cdot 169 = 338
    .
  - x = 20
    :
    (20 - 23)^2 = (-3)^2 = 9
    ,
    3 \cdot 9 = 27
    .
  - x = 30
    :
    (30 - 23)^2 = 7^2 = 49
    ,
    5 \cdot 49 = 245
    .
Sum the weighted squared deviations:
- 338 + 27 + 245 = 610
  .
Calculate the variance:
- \sigma^2 = \frac{\sum f \cdot (x - \mu)^2}{n} = \frac{610}{10} = 61
  .
Calculate the standard deviation:
- \sigma = \sqrt{61} \approx 7.810
  .
State the answer: The standard deviation is approximately 7.81.

Detailed Explanation:

Frequency distribution: Each value’s contribution to the variance is weighted by its frequency.
Formula adaptation: The variance formula becomes
\sigma^2 = \frac{\sum f \cdot (x - \mu)^2}{\sum f}
.
Interpretation: The standard deviation of 7.81 indicates significant spread, as values range from 10 to 30.

Answer:

\sigma \approx 7.81

Problem 5: Comparing Standard Deviations

Problem: Two datasets have the following values:

Dataset A: 5, 5, 5, 5
Dataset B: 3, 4, 6, 7
Which dataset has a higher standard deviation?

Solution Procedure:

Dataset A:
- Mean:
  \mu = \frac{5 + 5 + 5 + 5}{4} = 5
  .
- Since all values equal the mean,
  (x_i - \mu)^2 = 0
  .
- Variance:
  \sigma^2 = \frac{0}{4} = 0
  .
- Standard deviation:
  \sigma = 0
  .
Dataset B:
- Mean:
  \mu = \frac{3 + 4 + 6 + 7}{4} = \frac{20}{4} = 5
  .
- Squared deviations:
  - (3 - 5)^2 = 4
  - (4 - 5)^2 = 1
  - (6 - 5)^2 = 1
  - (7 - 5)^2 = 4
- Sum:
  4 + 1 + 1 + 4 = 10
  .
- Variance:
  \sigma^2 = \frac{10}{4} = 2.5
  .
- Standard deviation:
  \sigma = \sqrt{2.5} \approx 1.581
  .
Compare:
0 < 1.581
, so Dataset B has a higher standard deviation.
State the answer: Dataset B has a higher standard deviation.

Detailed Explanation:

Intuition: Dataset A has no variability (all values are identical), so its standard deviation is 0. Dataset B has varied values, leading to a positive standard deviation.
Why compare? Standard deviation indicates dispersion. Dataset B’s spread (values from 3 to 7) is greater than Dataset A’s (constant at 5).

Answer: Dataset B.

Problem 6: Standard Deviation of Grouped Data

Problem: The following table represents the heights (in cm) of students in class intervals:

Class Interval	Frequency
150–160	4
160–170	6
170–180	2
Find the standard deviation.

Solution Procedure:

Find the midpoint of each class:
- 150–160:
  \frac{150 + 160}{2} = 155
- 160–170:
  \frac{160 + 170}{2} = 165
- 170–180:
  \frac{170 + 180}{2} = 175
Calculate the mean:
- Total frequency (( n )):
  4 + 6 + 2 = 12
  .
- Sum of
  f \cdot x
  :
  (155 \cdot 4) + (165 \cdot 6) + (175 \cdot 2) = 620 + 990 + 350 = 1960
  .
- Mean (
  \mu
  ):
  \frac{1960}{12} \approx 163.333
  .
Calculate the squared deviations:
- x = 155
  :
  (155 - 163.333)^2 \approx (-8.333)^2 \approx 69.444
  ,
  4 \cdot 69.444 \approx 277.776
  .
- x = 165
  :
  (165 - 163.333)^2 \approx (1.667)^2 \approx 2.778
  ,
  6 \cdot 2.778 \approx 16.668
  .
- x = 175
  :
  (175 - 163.333)^2 \approx (11.667)^2 \approx 136.111
  ,
  2 \cdot 136.111 \approx 272.222
  .
Sum the weighted squared deviations:
- 277.776 + 16.668 + 272.222 \approx 566.666
  .
Calculate the variance:
- \sigma^2 = \frac{566.666}{12} \approx 47.222
  .
Calculate the standard deviation:
- \sigma = \sqrt{47.222} \approx 6.872
  .
State the answer: The standard deviation is approximately 6.87.

Detailed Explanation:

Grouped data: Midpoints approximate the data within each interval.
Assumption: Each value in a class is represented by its midpoint, a standard practice for continuous data.
Precision: Calculations are approximate due to rounding and the midpoint assumption.

Answer:

\sigma \approx 6.87

Problem 7: Effect of Adding a Constant

Problem: A dataset has values 4, 6, 8, 10 with a standard deviation of 2.236. If 5 is added to each value, what is the new standard deviation?

Solution Procedure:

Understand the effect: Adding a constant to each data point shifts the mean but does not affect the spread.
New dataset:
4+5, 6+5, 8+5, 10+5 = 9, 11, 13, 15
.
Property of standard deviation: The standard deviation is invariant to translation (adding a constant).
State the answer: The standard deviation remains 2.236.

Detailed Explanation:

Why unchanged? Standard deviation measures the spread relative to the mean. Adding 5 shifts all values (and the mean) by 5, leaving deviations
(x_i - \mu)
unchanged.
Verification: Calculate for the new dataset:
- Mean:
  \frac{9 + 11 + 13 + 15}{4} = 12
  .
- Variance: Same as original (deviations are identical).
- Standard deviation:
  \sqrt{5} \approx 2.236
  .

Answer:

\sigma = 2.236

Problem 8: Effect of Multiplying by a Constant

Problem: A dataset has a standard deviation of 3. If each value is multiplied by 2, what is the new standard deviation?

Solution Procedure:

Understand the effect: Multiplying each data point by a constant scales the spread.
Property of standard deviation: If each value is multiplied by ( k ), the standard deviation is multiplied by ( |k| ).
Apply the property: Here,
k = 2
, so new standard deviation =
3 \cdot 2 = 6
.
State the answer: The new standard deviation is 6.

Detailed Explanation:

Why scaled? Multiplying values by 2 doubles the differences between values and the mean, scaling the standard deviation.
Mathematical proof: If
y_i = k x_i
, then
\sigma_y = \sqrt{\frac{\sum (k x_i - k \mu)^2}{n}} = |k| \cdot \sigma_x
.
Intuition: The spread is stretched by the factor of 2.

Answer:

\sigma = 6

Problem 9: Standard Deviation of a Binomial Distribution

Problem: A binomial experiment involves 10 trials with a success probability of 0.4. Find the standard deviation of the number of successes.

Solution Procedure:

Identify the formula: For a binomial distribution with ( n ) trials and success probability ( p ), the standard deviation is
\sigma = \sqrt{n p (1-p)}
.
Plug in values:
n = 10
,
p = 0.4
,
1-p = 0.6
.
Calculate:
- n p (1-p) = 10 \cdot 0.4 \cdot 0.6 = 10 \cdot 0.24 = 2.4
  .
- \sigma = \sqrt{2.4} \approx 1.549
  .
State the answer: The standard deviation is approximately 1.55.

Detailed Explanation:

Binomial distribution: The number of successes follows a binomial distribution, where variance is
n p (1-p)
.
Why this formula? It accounts for the variability in the number of successes based on trial count and probability.
Interpretation: A standard deviation of 1.55 indicates the number of successes typically varies by about 1.55 from the mean (
n p = 4
).

Answer:

\sigma \approx 1.55

Problem 10: Combining Standard Deviations

Problem: Two independent datasets have the following properties:

Dataset A:
n_1 = 5
, mean = 10, standard deviation = 2.
Dataset B:
n_2 = 7
, mean = 15, standard deviation = 3.
Find the standard deviation of the combined dataset.

Solution Procedure:

Calculate the combined mean:
- Total sum:
  (5 \cdot 10) + (7 \cdot 15) = 50 + 105 = 155
  .
- Total ( n ):
  5 + 7 = 12
  .
- Mean:
  \frac{155}{12} \approx 12.917
  .
Use the formula for combined variance:
- Variance of combined dataset:
  \sigma^2 = \frac{n_1 \sigma_1^2 + n_2 \sigma_2^2 + n_1 (\mu_1 - \mu)^2 + n_2 (\mu_2 - \mu)^2}{n_1 + n_2}
  .
- \sigma_1^2 = 2^2 = 4
  ,
  \sigma_2^2 = 3^2 = 9
  .
- Deviations from combined mean:
  - \mu_1 - \mu = 10 - 12.917 \approx -2.917
    ,
    (-2.917)^2 \approx 8.509
    .
  - \mu_2 - \mu = 15 - 12.917 \approx 2.083
    ,
    (2.083)^2 \approx 4.339
    .
- Compute:
  n_1 \sigma_1^2 = 5 \cdot 4 = 20
  ,
  n_2 \sigma_2^2 = 7 \cdot 9 = 63
  .
- n_1 (\mu_1 - \mu)^2 = 5 \cdot 8.509 \approx 42.545
  ,
  n_2 (\mu_2 - \mu)^2 = 7 \cdot 4.339 \approx 30.373
  .
- Total:
  20 + 63 + 42.545 + 30.373 \approx 155.918
  .
- Variance:
  \sigma^2 = \frac{155.918}{12} \approx 12.993
  .
Calculate the standard deviation:
- \sigma = \sqrt{12.993} \approx 3.605
  .
State the answer: The standard deviation is approximately 3.61.

Detailed Explanation:

Combining datasets: The formula accounts for both the internal variability (
\sigma_1^2, \sigma_2^2
) and the variability due to differences in means.
Complexity: The additional terms adjust for the shift in the overall mean when datasets are merged.
Interpretation: The combined standard deviation reflects the increased spread due to differing means.

Answer:

\sigma \approx 3.61

Summary of Key Concepts

Standard deviation: Measures data dispersion around the mean.
Population vs. sample: Use ( n ) for population variance,
n-1
for sample variance.
Computational formula: Simplifies calculations for large datasets.
Frequency and grouped data: Weight deviations by frequency or use midpoints.
Transformations: Adding a constant leaves standard deviation unchanged; multiplying scales it.
Distributions: Binomial standard deviation uses
\sqrt{n p (1-p)}
.
Combining datasets: Accounts for both internal and between-group variability.

These problems cover a range of standard deviation applications, with detailed steps to ensure clarity.

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37. Standard deviation problems

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