Measures of Dispersion

23 soru

Soru 1Soru

What is the mean deviation of the data set 4,7,8,11,154, 7, 8, 11, 15?

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Cevap: 3.2

Cevap

The mean deviation of the data set is 3.2.
To find the mean deviation, first calculate the mean of the dataset, which is 9. Then, compute the absolute difference of each number from 9, obtaining values of 5, 2, 1, 2, and 6. Finally, divide the sum of these absolute values (16) by the total number of items (5) to get 3.2.

Adım Adım Çözüm

1
Calculate the arithmetic mean of the given data set.
The mean xˉ=9\bar{x} = 9.
The mean is needed to evaluate how far each data point deviates from the central value.
2
Find the absolute difference between each value and the mean.
The absolute deviations are 5, 2, 1, 2, and 6.
Mean deviation measures dispersion using absolute distances, ignoring negative signs.
3
Calculate the mean of the absolute deviations.
Mean deviation = 3.2.
Dividing the total sum of absolute deviations (16) by the number of observations (5) gives the mean deviation.

Anahtar Kavram

Mean Deviation for Ungrouped Data
Soru 2Soru

The set of numbers k2k - 2, kk, k+1k + 1, and k+5k + 5 is given, where kk is any real constant. What is the variance of this set of numbers?

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Cevap: 6.5

Cevap

The variance of the given set of numbers is 6.5.
The mean of the set is xˉ=k+1\bar{x} = k + 1. Subtracting the mean from each data point gives deviations of 3-3, 1-1, 00, and 44. The squares of these deviations are 99, 11, 00, and 1616, which sum to 2626. Dividing this sum by 44 gives a variance of 6.56.5. A key statistical property illustrated here is that adding or subtracting a constant kk from every value in a dataset shifts the mean by kk but leaves measures of dispersion (such as variance and standard deviation) unchanged.

Adım Adım Çözüm

1
Find the mean (\bar{x}) of the given set {k - 2, k, k + 1, k + 5}.
\bar{x} = \frac{(k - 2) + k + (k + 1) + (k + 5)}{4} = \frac{4k + 4}{4} = k + 1
The mean is calculated by summing all values and dividing by the total count of numbers.
2
Determine the deviation of each value from the mean, (x_i - \bar{x}).
(k - 2) - (k + 1) = -3, k - (k + 1) = -1, (k + 1) - (k + 1) = 0, (k + 5) - (k + 1) = 4
Deviations measure how far each data value lies from the mean.
3
Square each individual deviation and sum the results.
(-3)^2 + (-1)^2 + 0^2 + 4^2 = 9 + 1 + 0 + 16 = 26
Squaring converts all deviations into non-negative values.
4
Divide the sum of squared deviations by the total number of observations (N = 4) to find the variance.
Variance=264=6.5\text{Variance} = \frac{26}{4} = 6.5
Variance is defined as the arithmetic mean of the squared deviations from the mean.

Anahtar Kavram

Variance and Invariance under Constant Translation
Soru 3Soru

The table below shows the frequency distribution of marks obtained by a group of students in a mathematics test:

Mark (xx)246810
Frequency (ff)21412

Find the mean deviation of the distribution.

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Cevap: 2

Cevap

The mean deviation of the distribution is 2.
To find the mean deviation, first calculate the mean xˉ=fxf=6010=6\bar{x} = \frac{\sum fx}{\sum f} = \frac{60}{10} = 6. Next, sum the absolute deviations multiplied by their frequencies: fxxˉ=2(4)+1(2)+4(0)+1(2)+2(4)=20\sum f|x - \bar{x}| = 2(4) + 1(2) + 4(0) + 1(2) + 2(4) = 20. Dividing this total by the sum of frequencies 1010 yields a mean deviation of 22.

Adım Adım Çözüm

1
Calculate the arithmetic mean of the distribution
\bar{x} = \frac{\sum f x}{\sum f} = \frac{(2 \times 2) + (1 \times 4) + (4 \times 6) + (1 \times 8) + (2 \times 10)}{2 + 1 + 4 + 1 + 2} = \frac{60}{10} = 6
The mean is required as the central benchmark from which individual deviations are measured.
2
Calculate the sum of absolute deviations weighted by frequency
\sum f |x - \bar{x}| = 2|2 - 6| + 1|4 - 6| + 4|6 - 6| + 1|8 - 6| + 2|10 - 6| = 8 + 2 + 0 + 2 + 8 = 20
Each absolute difference from the mean must be multiplied by its frequency to account for the total deviation.
3
Divide the total absolute deviation by the total frequency
\text{Mean Deviation} = \frac{\sum f |x - \bar{x}|}{\sum f} = \frac{20}{10} = 2
The mean deviation represents the average distance of all observations from the arithmetic mean.

Anahtar Kavram

Mean Deviation of a Frequency Distribution
Tahmini Süre:1m 30s
Soru 4Soru

The test scores of five students in a mathematics quiz are 5,8,11,12,5, 8, 11, 12, and 1414. What is the variance of these test scores?

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Cevap: 1010

Cevap

The variance of the test scores is 1010.
The mean of the scores is 1010. Subtracting 1010 from each score gives deviations of 5,2,1,2,-5, -2, 1, 2, and 44. Squaring these gives 25,4,1,4,25, 4, 1, 4, and 1616, which sum to 5050. Dividing this sum by the number of data points (55) yields a variance of 1010.

Adım Adım Çözüm

1
Calculate the mean (xˉ\bar{x}) of the data set
xˉ=5+8+11+12+145=505=10\bar{x} = \frac{5 + 8 + 11 + 12 + 14}{5} = \frac{50}{5} = 10
The mean is required to find the deviation of each score from the central value.
2
Compute the squared deviations from the mean (xixˉ)2(x_i - \bar{x})^2
(510)2=25(5-10)^2 = 25, (810)2=4(8-10)^2 = 4, (1110)2=1(11-10)^2 = 1, (1210)2=4(12-10)^2 = 4, (1410)2=16(14-10)^2 = 16
Squaring deviations ensures all negative differences become positive.
3
Find the sum of all squared deviations
\sum (x_i - \bar{x})^2 = 25 + 4 + 1 + 4 + 16 = 50
Summing the squared deviations measures total variation around the mean.
4
Divide the total sum of squared deviations by the number of data values (N=5N = 5)
\text{Variance } (\sigma^2) = \frac{50}{5} = 10
Variance is defined as the average of the squared deviations.

Anahtar Kavram

Variance of Ungrouped Data
Tahmini Süre:1m 30s
Soru 5Soru

A small business recorded the number of customer inquiries received per day over six consecutive days as follows: 44, 77, 88, 1111, 1313, and 1717. What is the variance of the daily customer inquiries?

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Cevap: 18

Cevap

The variance of the daily customer inquiries is 1818.
To find the variance of the data set {4,7,8,11,13,17}\{4, 7, 8, 11, 13, 17\}, first calculate the mean: xˉ=4+7+8+11+13+176=10\bar{x} = \frac{4+7+8+11+13+17}{6} = 10. Next, compute the squared deviation of each data point from the mean: (410)2=36(4-10)^2 = 36, (710)2=9(7-10)^2 = 9, (810)2=4(8-10)^2 = 4, (1110)2=1(11-10)^2 = 1, (1310)2=9(13-10)^2 = 9, and (1710)2=49(17-10)^2 = 49. Summing these squared deviations gives 108108. Dividing this total by the number of observations (66) yields a variance of 1818.

Adım Adım Çözüm

1
Calculate the arithmetic mean of the given data set.
xˉ=10\bar{x} = 10
The mean is required as the central point from which deviations are calculated.
2
Determine the squared deviation of each data value from the mean.
(6)2=36(-6)^2 = 36, (3)2=9(-3)^2 = 9, (2)2=4(-2)^2 = 4, 12=11^2 = 1, 32=93^2 = 9, 72=497^2 = 49
Variance measures the average squared distance of data points from the mean.
3
Sum all calculated squared deviations.
(xxˉ)2=36+9+4+1+9+49=108\sum (x - \bar{x})^2 = 36 + 9 + 4 + 1 + 9 + 49 = 108
This provides the total sum of squares for the data set.
4
Divide the total sum of squares by the number of observations (n=6n = 6).
σ2=1086=18\sigma^2 = \frac{108}{6} = 18
Population variance formula is σ2=(xxˉ)2n\sigma^2 = \frac{\sum (x - \bar{x})^2}{n}.

Anahtar Kavram

Variance of Ungrouped Data
Tahmini Süre:1m 30s
Soru 6Soru

Five daily rainfall measurements (in mm) recorded in a city are 3,6,7,9,3, 6, 7, 9, and 1515. What is the standard deviation of these rainfall measurements?

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Cevap: 44

Cevap

The standard deviation of the rainfall measurements is 44.
First compute the mean of the data set: xˉ=3+6+7+9+155=8\bar{x} = \frac{3 + 6 + 7 + 9 + 15}{5} = 8. Next, sum the squared deviations from the mean: (38)2+(68)2+(78)2+(98)2+(158)2=25+4+1+1+49=80(3-8)^2 + (6-8)^2 + (7-8)^2 + (9-8)^2 + (15-8)^2 = 25 + 4 + 1 + 1 + 49 = 80. Dividing by the number of observations (55) yields the variance σ2=805=16\sigma^2 = \frac{80}{5} = 16. Taking the square root gives the standard deviation σ=16=4\sigma = \sqrt{16} = 4.

Adım Adım Çözüm

1
Calculate the arithmetic mean (xˉ\bar{x}) of the dataset
xˉ=3+6+7+9+155=405=8\bar{x} = \frac{3 + 6 + 7 + 9 + 15}{5} = \frac{40}{5} = 8
The mean is needed to find deviations for each data point.
2
Find the deviation of each number from the mean and square it
(38)2=25(3-8)^2 = 25, (68)2=4(6-8)^2 = 4, (78)2=1(7-8)^2 = 1, (98)2=1(9-8)^2 = 1, (158)2=49(15-8)^2 = 49
Squaring ensures all deviation values are non-negative.
3
Sum the squared deviations and divide by the total number of items (n=5n = 5) to find the variance
\text{Variance } (\sigma^2) = \frac{25 + 4 + 1 + 1 + 49}{5} = \frac{80}{5} = 16
Variance measures the average squared distance from the mean.
4
Take the square root of the variance to obtain standard deviation
\text{Standard Deviation } (\sigma) = \sqrt{16} = 4
Standard deviation returns the dispersion measure back to the original units.

Anahtar Kavram

Standard Deviation of Ungrouped Data
Soru 7Soru

The masses, in kilograms, of five wooden crates stored in a warehouse are 33, 77, 88, 1010, and 1212. What is the variance of the masses of these crates?

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Cevap: 9.29.2

Cevap

The variance of the masses is 9.29.2.
The mean of the numbers is 88. The squared deviations are 25,1,0,4,25, 1, 0, 4, and 1616. The sum of these squared deviations is 4646. Dividing by the count of numbers (55) gives 9.29.2.

Adım Adım Çözüm

1
Calculate the mean (xˉ\bar{x}) of the dataset.
xˉ=3+7+8+10+125=405=8\bar{x} = \frac{3 + 7 + 8 + 10 + 12}{5} = \frac{40}{5} = 8
The mean is required to find individual deviations.
2
Compute the deviation of each observation from the mean, (xxˉ)(x - \bar{x}).
38=53 - 8 = -5, 78=17 - 8 = -1, 88=08 - 8 = 0, 108=210 - 8 = 2, 128=412 - 8 = 4
Deviations quantify how far each data point lies from the central mean.
3
Square each deviation, (xxˉ)2(x - \bar{x})^2.
(5)2=25(-5)^2 = 25, (1)2=1(-1)^2 = 1, 02=00^2 = 0, 22=42^2 = 4, 42=164^2 = 16
Squaring eliminates negative signs and weights larger deviations.
4
Sum the squared deviations and divide by the total number of observations N=5N = 5.
Variance σ2=25+1+0+4+165=465=9.2\sigma^2 = \frac{25 + 1 + 0 + 4 + 16}{5} = \frac{46}{5} = 9.2
Variance is defined as the mean of the squared deviations.

Anahtar Kavram

Variance of Ungrouped Data
Tahmini Süre:1m 30s
Soru 8Soru

The arithmetic mean of five observations 11, 33, xx, 99, and 1212 is 66. What is the variance of this set of observations?

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Cevap: 1616

Cevap

The variance of the set of observations is 1616.
First, the missing value is found by equating the sum of the five numbers to 5×6=305 \times 6 = 30, which gives x=5x = 5. The data set is 1,3,5,9,121, 3, 5, 9, 12. The deviations from the mean 66 are 5,3,1,3,6-5, -3, -1, 3, 6, and their squares are 25,9,1,9,3625, 9, 1, 9, 36. Summing these squared deviations gives 8080. Dividing by the number of observations (55) yields the variance of 1616.

Adım Adım Çözüm

1
Find the missing value xx using the arithmetic mean formula.
x=5x = 5
The mean of N=5N = 5 numbers is xˉ=6\bar{x} = 6, so the total sum is 5×6=305 \times 6 = 30. Summing the known numbers gives 1+3+x+9+12=30    25+x=30    x=51 + 3 + x + 9 + 12 = 30 \implies 25 + x = 30 \implies x = 5.
2
Calculate the deviations (xixˉ)(x_i - \bar{x}) for each data value from the mean xˉ=6\bar{x} = 6.
Deviations are 5,3,1,3,6-5, -3, -1, 3, 6
Subtracting 66 from each value (1,3,5,9,12)(1, 3, 5, 9, 12) gives 16=51-6=-5, 36=33-6=-3, 56=15-6=-1, 96=39-6=3, and 126=612-6=6.
3
Square each deviation.
Squared deviations are 25,9,1,9,3625, 9, 1, 9, 36
Squaring each deviation yields (5)2=25(-5)^2 = 25, (3)2=9(-3)^2 = 9, (1)2=1(-1)^2 = 1, 32=93^2 = 9, and 62=366^2 = 36.
4
Compute the population variance σ2=(xixˉ)2N\sigma^2 = \frac{\sum (x_i - \bar{x})^2}{N}.
σ2=16\sigma^2 = 16
The sum of squared deviations is 25+9+1+9+36=8025 + 9 + 1 + 9 + 36 = 80. Dividing by N=5N = 5 gives 805=16\frac{80}{5} = 16.

Anahtar Kavram

Variance of Ungrouped Data with a Missing Observation
Soru 9Soru

The daily temperature readings, in degrees Celsius, recorded over four consecutive days are 1010, 1212, 1414, and 1616. What is the mean deviation of these temperatures?

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Cevap: 2

Cevap

The mean deviation of the temperatures is 2 °C.
To find the mean deviation, first calculate the mean of the data values: (10 + 12 + 14 + 16) / 4 = 13. Next, find the absolute difference of each value from the mean: |10 - 13| = 3, |12 - 13| = 1, |14 - 13| = 1, and |16 - 13| = 3. Finally, average these absolute differences: (3 + 1 + 1 + 3) / 4 = 8 / 4 = 2.

Adım Adım Çözüm

1
Calculate the arithmetic mean (average) of the data set
\(\bar{x} = \frac{10 + 12 + 14 + 16}{4} = 13\)
Mean deviation measures dispersion relative to the mean, so the mean must be calculated first.
2
Compute the absolute deviation of each value from the mean
\(|10 - 13| = 3\), \(|12 - 13| = 1\), \(|14 - 13| = 1\), \(|16 - 13| = 3\)
Mean deviation requires non-negative distances of each observation from the mean.
3
Sum the absolute deviations and divide by the sample size (N = 4)
\(\text{Mean Deviation} = \frac{3 + 1 + 1 + 3}{4} = \frac{8}{4} = 2\)
The mean deviation is the average of the absolute deviations.

Anahtar Kavram

Mean Deviation for Ungrouped Data
Soru 10Soru

A football coach records the number of goals scored by a team in six consecutive matches as 1,3,4,6,7,1, 3, 4, 6, 7, and 99. What is the variance of the goals scored?

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Cevap: 7

Cevap

The variance of the goals scored is 7.
To find the variance of the goals scored, first calculate the mean: xˉ=1+3+4+6+7+96=306=5\bar{x} = \frac{1+3+4+6+7+9}{6} = \frac{30}{6} = 5. Next, determine the sum of the squared deviations from the mean: (15)2+(35)2+(45)2+(65)2+(75)2+(95)2=16+4+1+1+4+16=42(1-5)^2 + (3-5)^2 + (4-5)^2 + (6-5)^2 + (7-5)^2 + (9-5)^2 = 16 + 4 + 1 + 1 + 4 + 16 = 42. Dividing this total by the number of data values (N=6N = 6) yields 426=7\frac{42}{6} = 7.

Adım Adım Çözüm

1
Calculate the mean of the dataset
\bar{x} = 5
The mean is needed as the reference point for computing deviations.
2
Compute the squared deviation of each data point from the mean
Squared deviations are 16, 4, 1, 1, 4, and 16
Variance measures the average squared distance from the mean.
3
Sum the squared deviations and divide by the number of observations N = 6
Variance = 7
The formula for variance of ungrouped data is \sigma^2 = \frac{\sum (x - \bar{x})^2}{N}.

Anahtar Kavram

Variance of Ungrouped Data
Soru 11Soru

The heights (in cm) of a seedling recorded over five consecutive weeks are 99, 1313, 1515, 1717, and 2121. What is the standard deviation of the heights?

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Cevap: 4

Cevap

The standard deviation of the seedling heights is 4 cm.
To find the standard deviation, first compute the mean: (9 + 13 + 15 + 17 + 21) / 5 = 15. Next, calculate the sum of squared deviations: (-6)^2 + (-2)^2 + 0^2 + 2^2 + 6^2 = 36 + 4 + 0 + 4 + 36 = 80. Divide by 5 to find the variance of 16. Finally, taking the square root of 16 gives the standard deviation of 4 cm.

Adım Adım Çözüm

1
Calculate the mean of the data values
Mean = 15 cm
The mean is necessary to calculate the deviations of each data point.
2
Calculate the squared deviations from the mean
Squared deviations are 36, 4, 0, 4, and 36 (sum = 80)
Variance measures the average of squared deviations from the mean.
3
Calculate the population variance
Variance = 80 / 5 = 16
Dividing the sum of squared deviations by N gives the variance.
4
Calculate the standard deviation
Standard deviation = sqrt(16) = 4 cm
The standard deviation is the square root of the variance.

Anahtar Kavram

Standard Deviation of Ungrouped Data
Soru 12Soru

The table below shows the frequency distribution of marks obtained by a group of students in a short test:

Mark (xx)1234
Frequency (ff)2332

What is the mean deviation of the distribution?

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Cevap: 0.90.9

Cevap

The mean deviation of the distribution is 0.90.9.
The mean of the data is xˉ=2510=2.5\bar{x} = \frac{25}{10} = 2.5. Calculating the sum of weighted absolute deviations yields 2(1.5)+3(0.5)+3(0.5)+2(1.5)=9.02(1.5) + 3(0.5) + 3(0.5) + 2(1.5) = 9.0. Dividing by total frequency 1010 gives the mean deviation as 0.90.9.

Adım Adım Çözüm

1
Calculate the mean (xˉ\bar{x}) of the frequency distribution.
f=2+3+3+2=10\sum f = 2 + 3 + 3 + 2 = 10, fx=(2×1)+(3×2)+(3×3)+(2×4)=2+6+9+8=25\sum fx = (2 \times 1) + (3 \times 2) + (3 \times 3) + (2 \times 4) = 2 + 6 + 9 + 8 = 25. Therefore, xˉ=2510=2.5\bar{x} = \frac{25}{10} = 2.5.
The mean is required to determine the deviations of each score value.
2
Find the absolute deviation xxˉ|x - \bar{x}| for each score value.
For x=1x = 1: 12.5=1.5|1 - 2.5| = 1.5; for x=2x = 2: 22.5=0.5|2 - 2.5| = 0.5; for x=3x = 3: 32.5=0.5|3 - 2.5| = 0.5; for x=4x = 4: 42.5=1.5|4 - 2.5| = 1.5.
Mean deviation measures average distance from the mean, ignoring signs.
3
Multiply each absolute deviation by its corresponding frequency and sum them.
fxxˉ=(2×1.5)+(3×0.5)+(3×0.5)+(2×1.5)=3+1.5+1.5+3=9.0\sum f|x - \bar{x}| = (2 \times 1.5) + (3 \times 0.5) + (3 \times 0.5) + (2 \times 1.5) = 3 + 1.5 + 1.5 + 3 = 9.0.
Frequencies reflect how many times each deviation occurs in the dataset.
4
Compute the mean deviation by dividing the weighted sum by the total frequency f\sum f.
\text{Mean Deviation} = \frac{\sum f|x - \bar{x}|}{\sum f} = \frac{9.0}{10} = 0.9.
The mean deviation is the average of these absolute deviations across all observations.

Anahtar Kavram

Mean Deviation for Discrete Frequency Distribution
Soru 13Soru

The ages, in years, of five participants in a workshop are 2,4,5,7,2, 4, 5, 7, and 1212. What is the mean deviation of the ages?

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Cevap: 2.82.8

Cevap

2.82.8
The mean deviation is calculated by taking the average of the absolute differences between each data point and the mean. For the dataset 2,4,5,7,2, 4, 5, 7, and 1212, the mean is 66. The absolute deviations from 66 are 4,2,1,1,4, 2, 1, 1, and 66, which sum to 1414. Dividing 1414 by 55 gives 2.82.8.

Adım Adım Çözüm

1
Calculate the arithmetic mean (xˉ\bar{x}) of the dataset.
xˉ=2+4+5+7+125=305=6\bar{x} = \frac{2 + 4 + 5 + 7 + 12}{5} = \frac{30}{5} = 6
The mean is needed as the reference point to calculate deviations.
2
Calculate the absolute deviation xxˉ|x - \bar{x}| for each data item.
26=4,46=2,56=1,76=1,126=6|2 - 6| = 4, \quad |4 - 6| = 2, \quad |5 - 6| = 1, \quad |7 - 6| = 1, \quad |12 - 6| = 6
Mean deviation measures distance from the mean, so absolute values are taken.
3
Find the average of these absolute deviations.
\text{Mean Deviation} = \frac{4 + 2 + 1 + 1 + 6}{5} = \frac{14}{5} = 2.8
The formula for mean deviation of ungrouped data is xxˉn\frac{\sum |x - \bar{x}|}{n}.

Anahtar Kavram

Mean Deviation for Ungrouped Data
Soru 14Soru

The test scores of six students in a mathematics quiz are 1212, 1818, 2525, 3131, 4747, and 5050. What is the range of these scores?

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Cevap: 3838

Cevap

The range of the scores is 3838.
The range is a measure of dispersion defined as the difference between the maximum value (5050) and the minimum value (1212). Subtracting 1212 from 5050 gives 3838.

Adım Adım Çözüm

1
Identify the maximum and minimum values in the dataset.
Maximum score = 5050, Minimum score = 1212.
Range measures the spread between the highest and lowest values in a distribution.
2
Calculate the range using the formula Range=MaximumMinimum\text{Range} = \text{Maximum} - \text{Minimum}.
Range=5012=38\text{Range} = 50 - 12 = 38.
Subtracting the minimum value from the maximum value yields the measure of dispersion known as range.

Anahtar Kavram

Range of a Dataset
Soru 15Soru

A library recorded the number of books borrowed over five consecutive days as 22, 44, 55, 66, and 88. What is the standard deviation of the number of books borrowed?

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Cevap: 22

Cevap

The standard deviation of the number of books borrowed is 22.
The mean of the five data points is 55. The sum of the squared deviations from the mean is 9+1+0+1+9=209 + 1 + 0 + 1 + 9 = 20. Dividing by 55 gives a variance of 44. Taking the square root of 44 yields the standard deviation of 22.

Adım Adım Çözüm

1
Calculate the arithmetic mean (xˉ\bar{x}) of the given dataset.
xˉ=2+4+5+6+85=255=5\bar{x} = \frac{2 + 4 + 5 + 6 + 8}{5} = \frac{25}{5} = 5
The mean is required to determine the deviations of each data point.
2
Find the deviations from the mean (xxˉ)(x - \bar{x}) and square each deviation (xxˉ)2(x - \bar{x})^2.
(25)2=9,(45)2=1,(55)2=0,(65)2=1,(85)2=9(2-5)^2 = 9, \quad (4-5)^2 = 1, \quad (5-5)^2 = 0, \quad (6-5)^2 = 1, \quad (8-5)^2 = 9
Squaring ensures all deviation values are positive before summation.
3
Calculate the variance (mean of squared deviations).
Variance (σ2)=9+1+0+1+95=205=4\text{Variance } (\sigma^2) = \frac{9 + 1 + 0 + 1 + 9}{5} = \frac{20}{5} = 4
Variance measures the average squared spread around the mean.
4
Take the square root of the variance to find the standard deviation.
Standard Deviation (σ)=4=2\text{Standard Deviation } (\sigma) = \sqrt{4} = 2
Standard deviation expresses dispersion in the original unit of measurement.

Anahtar Kavram

Standard Deviation of Ungrouped Data
Soru 16Soru

The set of four numbers {x,6,8,10}\{x, 6, 8, 10\} has a variance of 55. Given that x<6x < 6, find the value of xx.

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Cevap: 4

Cevap

The value of xx is 4.
By expressing the mean as xˉ=x+244\bar{x} = \frac{x+24}{4} and setting the sum of squared deviations divided by 4 equal to 5, we arrive at the quadratic equation x216x+48=0x^2 - 16x + 48 = 0. Solving this equation gives two possible values, x=4x = 4 and x=12x = 12. Applying the restriction x<6x < 6 uniquely identifies x=4x = 4 as the correct solution.

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1
Calculate the mean of the dataset in terms of xx
xˉ=x+244\bar{x} = \frac{x + 24}{4}
The mean of a dataset is the sum of all values divided by the total number of items.
2
Write the variance equation using the formula σ2=(xixˉ)2N\sigma^2 = \frac{\sum (x_i - \bar{x})^2}{N}
\frac{(x - \bar{x})^2 + (6 - \bar{x})^2 + (8 - \bar{x})^2 + (10 - \bar{x})^2}{4} = 5
The given variance is 5 and the number of observations N=4N = 4.
3
Substitute xˉ=x+244\bar{x} = \frac{x + 24}{4} and expand the squared deviations
\frac{3}{16}x^2 - 3x + 14 = 5 \implies x^2 - 16x + 48 = 0
Expanding the squared terms and multiplying through by 163\frac{16}{3} yields a standard quadratic equation.
4
Solve the quadratic equation x216x+48=0x^2 - 16x + 48 = 0 for xx
(x - 4)(x - 12) = 0 \implies x = 4 \text{ or } x = 12
Factoring the quadratic equation gives two possible roots.
5
Apply the given condition x<6x < 6
x = 4
Since x<6x < 6, we reject x=12x = 12 and select x=4x = 4.

Anahtar Kavram

Variance of Ungrouped Data and Algebraic Problem Solving
Soru 17Soru

Five packages delivered by a courier service have masses of 5 kg5\text{ kg}, 8 kg8\text{ kg}, 11 kg11\text{ kg}, 12 kg12\text{ kg}, and 14 kg14\text{ kg}. What is the mean deviation of the masses of these packages?

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Cevap: 2.8 kg2.8\text{ kg}

Cevap

2.8 kg2.8\text{ kg}
The mean of the data set is 10 kg10\text{ kg}. The distances of each data value from the mean are 55, 22, 11, 22, and 44. The sum of these distances is 1414, and dividing by 55 yields a mean deviation of 2.8 kg2.8\text{ kg}.

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1
Calculate the arithmetic mean (xˉ)(\bar{x}) of the dataset
\bar{x} = \frac{5 + 8 + 11 + 12 + 14}{5} = \frac{50}{5} = 10\text{ kg}
Mean deviation requires the central mean value as a reference point for all deviations.
2
Find the absolute deviation xxˉ|x - \bar{x}| for each value in the dataset
|5 - 10| = 5, |8 - 10| = 2, |11 - 10| = 1, |12 - 10| = 2, |14 - 10| = 4
Mean deviation measures the average distance of values from the mean regardless of sign.
3
Sum the absolute deviations and divide by the sample size (n=5)(n = 5)
\text{Mean Deviation} = \frac{5 + 2 + 1 + 2 + 4}{5} = \frac{14}{5} = 2.8\text{ kg}
Dividing total absolute deviation by the total count yields the mean deviation.

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Mean Deviation
Soru 18Soru

A meteorologist recorded the rainfall (in millimeters) in a town over five days as 22, 55, 66, 77, and 1010. What is the mean deviation of the rainfall data?

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Cevap: 22

Cevap

The mean deviation of the rainfall data is 22.
To find the mean deviation, first calculate the mean of the data: (2+5+6+7+10)/5=6(2 + 5 + 6 + 7 + 10) / 5 = 6. Next, calculate the absolute difference of each value from 6, which yields 4, 1, 0, 1, and 4. The average of these absolute differences is (4+1+0+1+4)/5=10/5=2(4 + 1 + 0 + 1 + 4) / 5 = 10 / 5 = 2.

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1
Calculate the arithmetic mean (xˉ\bar{x}) of the dataset.
\bar{x} = \frac{2 + 5 + 6 + 7 + 10}{5} = \frac{30}{5} = 6
The mean is required to compute the deviation of each observation from the central value.
2
Find the absolute deviation xxˉ|x - \bar{x}| for each data point.
|2 - 6| = 4, \quad |5 - 6| = 1, \quad |6 - 6| = 0, \quad |7 - 6| = 1, \quad |10 - 6| = 4
Mean deviation measures the average absolute distance of data values from the mean.
3
Sum the absolute deviations and divide by the total number of observations (N=5N = 5).
\text{Mean Deviation} = \frac{4 + 1 + 0 + 1 + 4}{5} = \frac{10}{5} = 2
Dividing the sum of absolute deviations by NN gives the mean deviation.

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Mean Deviation
Soru 19Soru

The mean of five numbers 3,5,7,x,3, 5, 7, x, and yy is 66. Given that the variance of the numbers is 88 and x<yx < y, calculate the value of yy.

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Cevap: 11

Cevap

11
The total sum of the five numbers is 5×6=305 \times 6 = 30, giving x+y=15x + y = 15. The sum of squared deviations from the mean 66 is 5×8=405 \times 8 = 40. The known numbers 3,5,73, 5, 7 contribute (3)2+(1)2+12=11(-3)^2 + (-1)^2 + 1^2 = 11 to this sum, leaving (x6)2+(y6)2=29(x-6)^2 + (y-6)^2 = 29. Substituting y=15xy = 15 - x yields 2x230x+88=0    x215x+44=02x^2 - 30x + 88 = 0 \implies x^2 - 15x + 44 = 0. Factoring gives roots 44 and 1111. Since x<yx < y, we find y=11y = 11.

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1
Use the definition of the arithmetic mean to write a linear relationship between xx and yy.
x+y=15x + y = 15, or y=15xy = 15 - x.
The total sum of 5 numbers with a mean of 6 is 5×6=305 \times 6 = 30. Subtracting the known numbers 3+5+7=153 + 5 + 7 = 15 leaves x+y=15x + y = 15.
2
Apply the variance formula for population data.
(xi6)2=40\sum (x_i - 6)^2 = 40.
Variance is the mean of squared deviations from the mean: (xixˉ)25=8    (xi6)2=40\frac{\sum (x_i - \bar{x})^2}{5} = 8 \implies \sum (x_i - 6)^2 = 40.
3
Compute the sum of squared deviations for the known elements and simplify the variance equation.
(x6)2+(y6)2=29(x - 6)^2 + (y - 6)^2 = 29.
The squared deviations for 3,5,73, 5, 7 are (3)2=9(-3)^2 = 9, (1)2=1(-1)^2 = 1, and 12=11^2 = 1. Subtracting 9+1+1=119 + 1 + 1 = 11 from 4040 gives 2929.
4
Substitute y=15xy = 15 - x into the simplified equation and solve the resulting quadratic equation.
x=4x = 4 or x=11x = 11.
Substituting y=15xy = 15 - x gives (x6)2+(9x)2=29    2x230x+88=0    x215x+44=0    (x4)(x11)=0(x - 6)^2 + (9 - x)^2 = 29 \implies 2x^2 - 30x + 88 = 0 \implies x^2 - 15x + 44 = 0 \implies (x - 4)(x - 11) = 0.
5
Select the correct pair (x,y)(x, y) using the condition x<yx < y.
x=4x = 4 and y=11y = 11.
Since x<yx < y, xx must be the smaller value (44) and yy must be the larger value (1111).

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Calculation of variance and mean for ungrouped data containing unknown elements
Soru 20Soru

The table below shows the distribution of scores obtained by a group of candidates in an examination:

Score (xx)246810
Frequency (ff)12621

Calculate the variance of the scores.

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Cevap: 4

Cevap

The variance of the scores is 4.
The mean score is xˉ=6\bar{x} = 6. Summing the weighted squared deviations gives f(x6)2=48\sum f(x - 6)^2 = 48. Dividing by the total frequency N=12N = 12 results in a variance of 4812=4\frac{48}{12} = 4.

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1
Find the total frequency and calculate the mean score.
Total frequency N=12N = 12 and mean xˉ=6\bar{x} = 6.
The mean is given by xˉ=fxf=2+8+36+16+1012=7212=6\bar{x} = \frac{\sum f x}{\sum f} = \frac{2 + 8 + 36 + 16 + 10}{12} = \frac{72}{12} = 6.
2
Determine the sum of squared deviations multiplied by their frequencies.
\sum f(x - \bar{x})^2 = 48.
Evaluating each term: 1(26)2=161(2-6)^2 = 16, 2(46)2=82(4-6)^2 = 8, 6(66)2=06(6-6)^2 = 0, 2(86)2=82(8-6)^2 = 8, and 1(106)2=161(10-6)^2 = 16. Summing these yields 16+8+0+8+16=4816 + 8 + 0 + 8 + 16 = 48.
3
Compute the variance by dividing the total squared deviations by the total frequency.
Variance = 4.
Variance is σ2=f(xxˉ)2f=4812=4\sigma^2 = \frac{\sum f(x - \bar{x})^2}{\sum f} = \frac{48}{12} = 4.

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Variance of a Frequency Distribution
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