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Assam Higher Secondary Statistics study materials
Assam Higher Secondary Statistics study resources

Assam Higher Secondary Statistics practice questions

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Question 1Descriptive Statistics: Collection and Presentation of Data, Measures of Central Tendency

Data collected by an investigator directly from the field for the first time, for the specific enquiry at hand, are called:

  1. Continuous data
  2. Secondary data
  3. Primary data Correct answer.
  4. Published data

Explanation

Primary data are those collected first-hand by the investigator for the enquiry being conducted, through surveys, experiments or direct observation. Data already gathered and issued by another agency are secondary, so they do not fit this definition.

Question 2Descriptive Statistics: Collection and Presentation of Data, Measures of Central Tendency

An ogive in statistics is best described as:

  1. A cumulative frequency curve Correct answer.
  2. A histogram of the data
  3. A frequency polygon of the data
  4. A pie diagram of the data

Explanation

An ogive, also called a cumulative frequency curve, is drawn by plotting cumulative frequencies against class boundaries. It is used to locate the median and quartiles graphically.

Question 3Descriptive Statistics: Collection and Presentation of Data, Measures of Central Tendency

The mode of a distribution is the value that:

  1. is the middle-most value after arrangement
  2. occurs most frequently Correct answer.
  3. divides the distribution into two equal parts
  4. is the average of all observations

Explanation

The mode is the observation appearing with the highest frequency in the data set. It is a positional measure of central tendency and is not affected by extreme values.

Question 4Descriptive Statistics: Collection and Presentation of Data, Measures of Central Tendency

The mean of 5 observations is 12. If one observation, 15, is replaced by 25, the new mean is:

  1. 12
  2. 12.5
  3. 14 Correct answer.
  4. 13

Explanation

The original total of the five observations is 5 × 12 = 60. Replacing 15 by 25 raises the total by 10, giving 70, so the new mean is 70 ÷ 5 = 14.

Question 5Descriptive Statistics: Collection and Presentation of Data, Measures of Central Tendency

The mean marks of 30 boys is 60 and the mean marks of 20 girls is 70. The mean marks of the combined group of 50 students is:

  1. 66
  2. 68
  3. 64 Correct answer.
  4. 65

Explanation

The combined mean is the total marks of both groups divided by the total number of students: (30×60 + 20×70) ÷ 50 = 3200 ÷ 50 = 64. A simple average of 60 and 70 would wrongly ignore the unequal group sizes.

Question 6Descriptive Statistics: Collection and Presentation of Data, Measures of Central Tendency

The median of the observations 7, 3, 9, 1, 5 is:

  1. 5 Correct answer.
  2. 7
  3. 3
  4. 9

Explanation

Arranging the observations in ascending order gives 1, 3, 5, 7, 9. Since there are five observations, the median is the third value, which is 5.

Question 7Descriptive Statistics: Collection and Presentation of Data, Measures of Central Tendency

For the observations 4, 8, 10, 12, 16, 20, 24, the first quartile Q1 is:

  1. 8 Correct answer.
  2. 7
  3. 10
  4. 6

Explanation

For n = 7, the first quartile is the value at position (n+1)/4 = 2 in the ordered data. The second observation in 4, 8, 10, 12, 16, 20, 24 is 8.

Question 8Descriptive Statistics: Collection and Presentation of Data, Measures of Central Tendency

For a moderately skewed distribution the empirical relation is Mode = 3 Median − 2 Mean. If the mean is 30 and the mode is 24, the median is:

  1. 29
  2. 28 Correct answer.
  3. 27
  4. 26

Explanation

Substituting into Mode = 3 Median − 2 Mean gives 24 = 3M − 60, so 3M = 84 and M = 28. The relation is approximate and is applied only to moderately skewed distributions.

Question 9Descriptive Statistics: Collection and Presentation of Data, Measures of Central Tendency

A student scores 60, 70 and 80 marks in three subjects, carrying weights 2, 3 and 5 respectively. The weighted mean mark is:

  1. 70
  2. 73 Correct answer.
  3. 72
  4. 75

Explanation

The weighted mean is Σwx ÷ Σw = (2×60 + 3×70 + 5×80) ÷ (2 + 3 + 5) = 730 ÷ 10 = 73. The weights reflect the relative importance of each subject.

Question 10Descriptive Statistics: Collection and Presentation of Data, Measures of Central Tendency

In a histogram, the area of each rectangle is proportional to:

  1. the class width of the class
  2. the cumulative frequency of the class
  3. the class mark of the class
  4. the frequency of the class Correct answer.

Explanation

In a histogram the area of each rectangle, not its height alone, is proportional to the class frequency. When class widths are equal, the heights themselves are also proportional to the frequencies.

Question 11Descriptive Statistics: Collection and Presentation of Data, Measures of Central Tendency

A municipal office prepares a table of household incomes in a town using entries already recorded in its own tax register. For the investigator using this table, the data are best classified as:

  1. Nominal data
  2. Secondary data Correct answer.
  3. Primary data
  4. Qualitative data

Explanation

Data already collected by another agency or taken from an existing record and used by a later investigator are secondary data. Because the municipal tax register was not compiled by the present investigator for this specific enquiry, the table is a secondary source.

Question 12Descriptive Statistics: Collection and Presentation of Data, Measures of Central Tendency

For the class interval 30–40 in a frequency table, the class mark used in computing the arithmetic mean is:

  1. 30
  2. 40
  3. 35 Correct answer.
  4. 70

Explanation

The class mark is the midpoint of a class interval, found as (lower limit + upper limit)/2. For 30–40 this is (30+40)/2 = 35, and this midpoint represents all observations in the class.

Question 13Descriptive Statistics: Collection and Presentation of Data, Measures of Central Tendency

In a pie diagram showing the expenditure of a family on food, rent, education and other items, the angle of each sector is taken proportional to:

  1. The magnitude of the corresponding component Correct answer.
  2. The cumulative frequencies of the items
  3. The class marks of the classes
  4. The midpoints of the classes

Explanation

A pie chart divides a circle into sectors so that each sector angle represents the share of that component in the total. The angle is calculated as (component value ÷ total value) × 360°, so it is proportional to the component magnitude.

Question 14Descriptive Statistics: Collection and Presentation of Data, Measures of Central Tendency

A researcher arranges all names in a college register alphabetically, selects one name at random from the first ten, and then takes every tenth name after it. The sampling method used is:

  1. Systematic sampling Correct answer.
  2. Simple random sampling
  3. Quota sampling
  4. Stratified random sampling

Explanation

Systematic sampling starts with a random item in the first interval and then selects every kth item, here every tenth name. This gives a fixed interval between selected units after the random start.

Question 15Descriptive Statistics: Collection and Presentation of Data, Measures of Central Tendency

In a grouped frequency distribution, the cumulative frequency below 20 is 20 and the cumulative frequency below 30 is 45. The frequency of the class 20–30 is:

  1. 30
  2. 25 Correct answer.
  3. 45
  4. 20

Explanation

The frequency of a class is the difference between the cumulative frequencies at its upper and lower class boundaries. Here it is 45 − 20 = 25, so the class 20–30 contains 25 observations.

Question 16Descriptive Statistics: Collection and Presentation of Data, Measures of Central Tendency

For the discrete distribution with values 2, 4, 6, 8 having frequencies 3, 2, 4, 1 respectively, the arithmetic mean is:

  1. 4.2
  2. 5.4
  3. 4.6 Correct answer.
  4. 5.0

Explanation

For a discrete frequency distribution, mean = Σfx ÷ Σf. Here Σfx = 2(3)+4(2)+6(4)+8(1)=46 and Σf=10, giving mean 4.6.

Question 17Descriptive Statistics: Collection and Presentation of Data, Measures of Central Tendency

For the distribution with classes 0–10, 10–20, 20–30, 30–40, 40–50 and frequencies 4, 6, 10, 8, 2, the median is:

  1. 22
  2. 25 Correct answer.
  3. 27
  4. 30

Explanation

Total frequency N=30, so the median item is the 15th observation. The cumulative frequencies are 4, 10, 20, 28, 30, placing the median in 20–30; median = 20 + [(15−10)/10]×10 = 25.

Question 18Descriptive Statistics: Collection and Presentation of Data, Measures of Central Tendency

The mean of the first three of five observations is 10, the mean of the last three is 12, and the third observation is 8. The mean of all five observations is:

  1. 10.8
  2. 11.6 Correct answer.
  3. 11.2
  4. 12.0

Explanation

The first three observations sum to 3×10=30 and the last three sum to 3×12=36. Adding these sums counts the third observation twice, so total sum of five = 30+36−8=58, and mean = 58/5 = 11.6.

Question 19Descriptive Statistics: Collection and Presentation of Data, Measures of Central Tendency

For the ordered observations 12, 15, 20, 25, 30, 35, 40, the third decile D3 is:

  1. 24
  2. 15
  3. 20
  4. 17 Correct answer.

Explanation

For n=7, the position of D3 is 3(n+1)/10 = 3×8/10 = 2.4. The second observation is 15 and the third is 20, so D3 = 15 + 0.4(20−15) = 17.

Question 20Descriptive Statistics: Collection and Presentation of Data, Measures of Central Tendency

In a frequency distribution with inclusive classes 10–19, 20–29, 30–39, the class boundaries of the class 20–29 are:

  1. 19–30
  2. 19.5–29.5 Correct answer.
  3. 20–29
  4. 20.5–29.5

Explanation

For inclusive classes, class boundaries are obtained by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit. Thus 20–29 becomes 19.5–29.5, removing the gap between consecutive classes.

Question 21Descriptive Statistics: Collection and Presentation of Data, Measures of Central Tendency

A researcher divides a college population into separate groups according to stream (Arts, Science and Commerce) and then draws a random sample from each group in proportion to its size. This sampling procedure is called:

  1. Quota sampling
  2. Stratified random sampling Correct answer.
  3. Systematic sampling
  4. Cluster sampling

Explanation

Stratified random sampling divides the population into homogeneous strata such as streams and then draws a random sample from every stratum, often in proportion to stratum size. Systematic sampling selects every kth unit after a random start, while cluster sampling selects entire groups. Quota sampling is a non-probability method based on fixed quotas.

Question 22Descriptive Statistics: Collection and Presentation of Data, Measures of Central Tendency

For a set of n observations with arithmetic mean x̄, the sum of the deviations (xi − x̄) is:

  1. Always positive
  2. Equal to the sum of the observations
  3. Zero Correct answer.
  4. Equal to n times the mean

Explanation

The arithmetic mean is the balance point of the data, so the positive and negative deviations cancel. Algebraically, Σ(xi − x̄) = Σxi − nx̄ = 0 because x̄ = Σxi/n. This holds for any set of observations, whether positive or negative.

Question 23Descriptive Statistics: Collection and Presentation of Data, Measures of Central Tendency

A frequency table that classifies data according to two or more characteristics simultaneously is known as:

  1. A manifold table Correct answer.
  2. A histogram
  3. An ogive
  4. A frequency polygon

Explanation

A table that simultaneously classifies data by two or more characteristics is called a manifold or multi-way table. Frequency polygons, ogives and histograms are graphical presentations, not tables. A simple table shows only one characteristic.

Question 24Descriptive Statistics: Collection and Presentation of Data, Measures of Central Tendency

For the frequency distribution with classes 0–10, 10–20, 20–30, 30–40 and 40–50 and frequencies 5, 15, 20, 8 and 2, the mean obtained by the step-deviation method with assumed mean 25 and class width 10 is:

  1. 24.0
  2. 22.4 Correct answer.
  3. 21.6
  4. 23.2

Explanation

Using midpoints 5, 15, 25, 35 and 45 and step deviations d = (x − 25)/10 gives −2, −1, 0, 1 and 2. Then Σfd = −13 and Σf = 50, so x̄ = 25 + 10(−13/50) = 22.4. The step-deviation method reduces arithmetic before applying the correction.

Question 25Descriptive Statistics: Collection and Presentation of Data, Measures of Central Tendency

For the discrete distribution with values 5, 10, 15, 20 and 25 having frequencies 4, 6, 10, 7 and 3 respectively, the median is:

  1. 10
  2. 15 Correct answer.
  3. 12.5
  4. 20

Explanation

The cumulative frequencies are 4, 10, 20, 27 and 30, so N/2 = 15. The first cumulative frequency reaching 15 occurs at x = 15, whose cumulative frequency is 20. Hence the median of this discrete distribution is 15.

Question 26Descriptive Statistics: Collection and Presentation of Data, Measures of Central Tendency

For the grouped frequency distribution with classes 10–20, 20–30, 30–40, 40–50 and 50–60 and frequencies 4, 8, 12, 6 and 2, the mode is:

  1. 32
  2. 36
  3. 35
  4. 34 Correct answer.

Explanation

The largest frequency is 12 in the class 30–40, so this is the modal class with l = 30, f1 = 12, f0 = 8, f2 = 6 and h = 10. Mode = 30 + [(12 − 8)/(2×12 − 8 − 6)]×10 = 30 + 4 = 34.

Question 27Descriptive Statistics: Collection and Presentation of Data, Measures of Central Tendency

The mean of the distribution with classes 0–10, 10–20, 20–30 and 30–40 and frequencies 5, f, 10 and 5 is 20. The missing frequency f is:

  1. 10 Correct answer.
  2. 8
  3. 6
  4. 12

Explanation

With midpoints 5, 15, 25 and 35, Σfx = 450 + 15f and Σf = 20 + f. Setting (450 + 15f)/(20 + f) = 20 gives 450 + 15f = 400 + 20f, so f = 10. This value also makes the total frequency 30.

Question 28Descriptive Statistics: Collection and Presentation of Data, Measures of Central Tendency

For a frequency distribution with open-end classes, which measure of central tendency is most appropriate?

  1. Arithmetic mean
  2. Weighted mean
  3. Geometric mean
  4. Median Correct answer.

Explanation

Open-end classes have no definite class limits, so their midpoints cannot be fixed and the arithmetic mean cannot be computed without assumptions. The median depends only on the position of the middle item and can be located from cumulative frequencies, so it is the suitable average. Geometric and weighted means also require numerical observations or weights.

Question 29Descriptive Statistics: Collection and Presentation of Data, Measures of Central Tendency

The median of the distribution with classes 0–10, 10–20, 20–30, 30–40 and 40–50 and frequencies 6, f, 15, 12 and 7 is 28. The missing frequency f is:

  1. 6
  2. 4 Correct answer.
  3. 5
  4. 3

Explanation

The total frequency is 40 + f, so N/2 = 20 + 0.5f. The median 28 lies in 20–30, where l = 20, cf = 6 + f and f_m = 15. Solving 20 + [((20 + 0.5f) − (6 + f))/15]×10 = 28 gives f = 4.

Question 30Descriptive Statistics: Collection and Presentation of Data, Measures of Central Tendency

The mean of 100 observations was calculated as 50. It was later found that two observations 60 and 70 were wrongly read as 16 and 7. The correct mean is:

  1. 50.93
  2. 51.30
  3. 51.07 Correct answer.
  4. 51.00

Explanation

The incorrect total is 100 × 50 = 5000. Replacing the wrong entries 16 and 7 by the correct 60 and 70 changes the total to 5000 − 16 − 7 + 60 + 70 = 5107. Hence the correct mean is 5107/100 = 51.07.

Question 31Descriptive Statistics: Collection and Presentation of Data, Measures of Central Tendency

In a questionnaire survey, a question that hints at the answer the investigator expects is known as:

  1. a pilot question
  2. a leading question Correct answer.
  3. an open question
  4. a closed question

Explanation

A leading question is worded so that it suggests or prompts a particular answer, which can bias survey results. A closed question gives fixed alternatives, an open question allows free response, and a pilot question is used in a trial run.

Question 32Descriptive Statistics: Collection and Presentation of Data, Measures of Central Tendency

The mode of the observations 4, 7, 7, 9, 10, 7, 12 is:

  1. 10
  2. 7 Correct answer.
  3. 9
  4. 4

Explanation

The mode is the observation with the highest frequency. In the given data, 7 occurs three times while every other value occurs once, so the mode is 7.

Question 33Descriptive Statistics: Collection and Presentation of Data, Measures of Central Tendency

Which measure of central tendency divides the ordered observations into two equal parts?

  1. Geometric mean
  2. Mode
  3. Median Correct answer.
  4. Arithmetic mean

Explanation

The median is a positional average. When the observations are arranged in order, it divides the series into two equal halves, so half the observations lie below it and half above it.

Question 34Descriptive Statistics: Collection and Presentation of Data, Measures of Central Tendency

In a histogram with unequal class widths, the height of the bar for the class 20–40 with frequency 30 is:

  1. 30
  2. 1.5 Correct answer.
  3. 10
  4. 20

Explanation

For a histogram with unequal class widths, the height of each bar is the frequency density, not the raw frequency. Here frequency density = frequency ÷ class width = 30 ÷ 20 = 1.5.

Question 35Descriptive Statistics: Collection and Presentation of Data, Measures of Central Tendency

The two ogives of a grouped frequency distribution meet at a point whose x-coordinate equals the:

  1. median Correct answer.
  2. range
  3. arithmetic mean
  4. mode

Explanation

The less-than and more-than ogives intersect at the cumulative frequency N/2. The x-coordinate of this intersection therefore gives the median of the grouped distribution.

Question 36Descriptive Statistics: Collection and Presentation of Data, Measures of Central Tendency

A candidate scores 80 marks in a test with weight 2 and 60 marks in a test with weight 3. The weighted arithmetic mean is:

  1. 72
  2. 70
  3. 64
  4. 68 Correct answer.

Explanation

The weighted arithmetic mean uses each value multiplied by its weight. It is (80×2 + 60×3)/(2+3) = (160+180)/5 = 68.

Question 37Descriptive Statistics: Collection and Presentation of Data, Measures of Central Tendency

The geometric mean of 4, 16 and 64 is:

  1. 12
  2. 28
  3. 8
  4. 16 Correct answer.

Explanation

The geometric mean of n positive values is the nth root of their product. Here GM = (4×16×64)^(1/3) = (4096)^(1/3) = 16.

Question 38Descriptive Statistics: Collection and Presentation of Data, Measures of Central Tendency

A car covers 60 km at 30 km/h and another 60 km at 60 km/h. The harmonic mean speed for the whole journey is:

  1. 36 km/h
  2. 40 km/h Correct answer.
  3. 48 km/h
  4. 45 km/h

Explanation

For equal distances travelled at different speeds, the harmonic mean gives the average speed. HM = 2/(1/30 + 1/60) = 40 km/h, which also follows from total distance 120 km in total time 3 h.

Question 39Descriptive Statistics: Collection and Presentation of Data, Measures of Central Tendency

For a moderately skewed distribution, mean = 30 and median = 32. Using the empirical relation, the mode is:

  1. 36 Correct answer.
  2. 34
  3. 30
  4. 32

Explanation

The empirical relation for a moderately skewed distribution is Mode = 3 Median − 2 Mean. Substituting the values gives 3(32) − 2(30) = 96 − 60 = 36.

Question 40Descriptive Statistics: Collection and Presentation of Data, Measures of Central Tendency

The mean of 20 observations is 15. If each observation is multiplied by 3 and then 2 is subtracted from each result, the mean of the new observations is:

  1. 41
  2. 43 Correct answer.
  3. 45
  4. 47

Explanation

The mean changes in the same linear way as each observation. If every observation is multiplied by 3 and then reduced by 2, the new mean is 3(15) − 2 = 43.

Question 41Measures of Dispersion, Moments, Skewness and Kurtosis

The range of the observations 12, 7, 19, 25, 4, 16 is:

  1. 18
  2. 12
  3. 21 Correct answer.
  4. 25

Explanation

The range is the difference between the largest and the smallest observation. Here the largest value is 25 and the smallest is 4, so the range is 25 − 4 = 21. The middle observations do not enter the calculation, which is why the range is a quick but crude measure of dispersion.

Question 42Measures of Dispersion, Moments, Skewness and Kurtosis

The quartile deviation of a distribution is given by:

  1. Q3 + Q1
  2. (Q3 + Q1)/2
  3. (Q3 − Q1)/2 Correct answer.
  4. Q3 − Q1

Explanation

The quartile deviation, also called the semi-interquartile range, is half the distance between the third and first quartiles, that is QD = (Q3 − Q1)/2. Because it depends only on the middle 50% of the observations, it is not affected by extreme values.

Question 43Measures of Dispersion, Moments, Skewness and Kurtosis

A frequency distribution has β2 = 2.1. It is best described as:

  1. Platykurtic, with a flatter peak Correct answer.
  2. Mesokurtic, like the normal curve
  3. Leptokurtic, with a sharper peak
  4. Symmetrical, with zero skewness

Explanation

Kurtosis is measured by β2, whose reference value for the normal curve is 3. A value of 2.1 is less than 3, so the distribution is platykurtic, meaning it is flatter than the normal curve with lighter tails. Symmetry is a separate property, judged from skewness and not from β2 alone.

Question 44Measures of Dispersion, Moments, Skewness and Kurtosis

The mean deviation about the mean of the observations 2, 4, 6, 8, 10 is:

  1. 2.4 Correct answer.
  2. 2.0
  3. 3.0
  4. 1.2

Explanation

The mean of the five observations is 6, so the absolute deviations from the mean are 4, 2, 0, 2 and 4. Their total is 12, and dividing by the number of observations, 5, gives a mean deviation about the mean of 2.4.

Question 45Measures of Dispersion, Moments, Skewness and Kurtosis

A data set has mean 50 and standard deviation 5. Its coefficient of variation is:

  1. 10% Correct answer.
  2. 5%
  3. 50%
  4. 0.1%

Explanation

The coefficient of variation expresses the standard deviation as a percentage of the mean: CV = (SD ÷ mean) × 100. Substituting gives (5 ÷ 50) × 100 = 10%, which indicates a comparatively consistent data set.

Question 46Measures of Dispersion, Moments, Skewness and Kurtosis

If every observation in a data set is multiplied by 3, the variance of the new data set will be:

  1. six times
  2. unchanged
  3. three times
  4. nine times Correct answer.

Explanation

Multiplying every observation by a constant k multiplies the mean and the standard deviation by k, and therefore multiplies the variance by k². With k = 3 the variance becomes 9 times its original value. The coefficient of variation, being a ratio of SD to mean, remains unchanged.

Question 47Measures of Dispersion, Moments, Skewness and Kurtosis

In terms of raw moments, the second central moment μ2 is equal to:

  1. μ2′ − (μ1′)² Correct answer.
  2. μ2′ − μ1′
  3. μ2′ + (μ1′)²
  4. (μ2′)² − μ1′

Explanation

The second central moment is the variance, and it is related to the raw moments by μ2 = μ2′ − (μ1′)². The subtracted term (μ1′)² corrects for the fact that deviations are taken about the mean rather than about zero.

Question 48Measures of Dispersion, Moments, Skewness and Kurtosis

For a distribution, Q1 = 20, median = 25 and Q3 = 40. Bowley's coefficient of skewness is:

  1. −0.5
  2. 0.25
  3. 0.5 Correct answer.
  4. 1.0

Explanation

Bowley's coefficient of skewness is Sk = (Q3 + Q1 − 2Q2) ÷ (Q3 − Q1). Substituting the given values gives (40 + 20 − 50) ÷ 20 = 10 ÷ 20 = 0.5, a small positive skew indicating the upper quartile lies farther from the median.

Question 49Measures of Dispersion, Moments, Skewness and Kurtosis

A distribution has mean 50, median 45 and standard deviation 10. Using the empirical relation between mean, median and mode, Karl Pearson's coefficient of skewness is:

  1. 1.5 Correct answer.
  2. 0.5
  3. 2.0
  4. 1.0

Explanation

When the mode is not given directly, the empirical relation Mode = 3 Median − 2 Mean gives Mode = 135 − 100 = 35. Karl Pearson's coefficient is then (Mean − Mode) ÷ SD = (50 − 35) ÷ 10 = 1.5, which equals 3(Mean − Median) ÷ SD as well.

Question 50Measures of Dispersion, Moments, Skewness and Kurtosis

For a distribution the first three raw moments about zero are μ1′ = 1, μ2′ = 4 and μ3′ = 10. The third central moment μ3 is:

  1. 0 Correct answer.
  2. −2
  3. 4
  4. 2

Explanation

The third central moment is obtained from raw moments by μ3 = μ3′ − 3μ2′μ1′ + 2(μ1′)³. Substituting gives 10 − 3(4)(1) + 2(1)³ = 10 − 12 + 2 = 0, so the distribution has no third-moment skewness, consistent with a symmetric shape.