Statistics MCQ Quiz - Objective Question with Answer for Statistics - Download Free PDF
Last updated on Apr 29, 2024
Latest Statistics MCQ Objective Questions
Statistics Question 1:
Infant mortality rate is the ratio of -
Answer (Detailed Solution Below)
Statistics Question 1 Detailed Solution
![](http://cdn.testbook.com/resources/lms_creative_elements/key-point-image.png)
- Infant mortality rate is the ratio of the number of deaths of infants (children under one year of age) during a given time period to the total number of live births during the same period.
- This rate is often expressed per 1,000 live births and is used as an important indicator of the overall health and well-being of a population.
- It provides insights into the quality of healthcare, nutrition, sanitation, and socioeconomic conditions, especially for newborns and infants.
Statistics Question 2:
Regression analysis in concerns with prediction of
Answer (Detailed Solution Below)
Statistics Question 2 Detailed Solution
The correct answer is "Dependent variable".
Key Points
- Regression Analysis-
- Regression analysis is a statistical technique for modeling and investigating the relationship between two or more variables, say a dependent variable, and one or more independent variables.
- The variable researchers are trying to explain or predict is called the response variable or, "dependent variable" because it depends on another variable.
- The earliest form of regression was the least squares method, published by Legendre in 1805, and by Gauss in 1809.
- The term "regression" was coined by Francis Galton in the 19th century to describe a biological phenomenon.
- Uses of Regression Analysis-
- Regression analysis is widely used for prediction and forecasting.
- In some situations, regression analysis can be used to infer causal relationships between the independent and dependent variables.
Additional Information
- Dependent Variable-
- Dependent variables depend, by some law or rule, on the values of other variables.
- Depending on the context, a dependent variable is sometimes called a "response variable", "regressand", "criterion", "predicted variable", "measured variable", "explained variable", "experimental variable", "responding variable", "outcome variable", "output variable", "target" or "label".
- Independent Variable-
- The independent variables are those that predict or forecast the values of the dependent variable in the model.
- Independent variables are not seen as depending on any other variable in the scope of the experiment.
- The term "predictor variable", "explanatory variable", 'predictors', 'covariates' or 'features' is used by authors for the independent variable.
Statistics Question 3:
Which of the following activities are 'economic'?
Answer (Detailed Solution Below)
Statistics Question 3 Detailed Solution
The correct answer is Joe has served 5 customers and they have paid Rs. 1200 for his service.
Key Points
- Economic Activity:
- Economic activity is occurred when a good or service is produced or consumed in order to earn some profit.
- For example, when a housewife is working full day but not getting paid, her work will not be considered as an economic activity.
- But, a teacher receives some amount of fees after one month of teaching would be considered under the category of economic activities.
Statistics Question 4:
If the difference between mean and mode is 48 and median is 12, Find mean.
Answer (Detailed Solution Below)
Statistics Question 4 Detailed Solution
Concept:
Mean - mode = 3(Mean - median)
Calculation:
Given:
The difference between mean and mode = 48
And median = 12
Mean - mode = 3(Mean - median)
⇒ 48 = 3(mean - 12)
⇒ 48 = 3 mean - 36
⇒ 3 mean = 48 + 36
⇒ 3 mean = 84
⇒ mean = 84/3
⇒ mean = 28
∴ Mean is 28
Statistics Question 5:
A simple random sample consists of three observations 1, 3, 5. The point estimate of the population standard deviation is :
Answer (Detailed Solution Below)
Statistics Question 5 Detailed Solution
Concept -
The formula for the sample standard deviation s is given by:
\(s = \sqrt{\frac{1}{n-1} \sum_{i=1}^n (x_i - \bar{x})^2}\)
Where:
- n is the number of observations in the sample.
- xi are the sample observations.
- \( \bar{x} \) is the sample mean.
Explanation -
First, calculate the sample mean \( \bar{x} \) :
\( \bar{x} \) = \(\frac{1+3+5}{3} = \frac{9}{3} = 3\)
Next, compute the sum of squared deviations from the mean:
\((1-3)^2 + (3-3)^2 + (5-3)^2 = (-2)^2 + 0^2 + 2^2 = 4 + 0 + 4 = 8\)
Now, use the formula for the sample standard deviation:
s = \(\sqrt{\frac{1}{3-1} \times 8} = \sqrt{\frac{8}{2}} = \sqrt{4} = 2\)
Therefore, the point estimate of the population standard deviation, based on the sample standard deviation is 2.
Top Statistics MCQ Objective Questions
What is the mean of the range, mode and median of the data given below?
5, 10, 3, 6, 4, 8, 9, 3, 15, 2, 9, 4, 19, 11, 4
Answer (Detailed Solution Below)
Statistics Question 6 Detailed Solution
Download Solution PDFGiven:
The given data is 5, 10, 3, 6, 4, 8, 9, 3, 15, 2, 9, 4, 19, 11, 4
Concept used:
The mode is the value that appears most frequently in a data set
At the time of finding Median
First, arrange the given data in the ascending order and then find the term
Formula used:
Mean = Sum of all the terms/Total number of terms
Median = {(n + 1)/2}th term when n is odd
Median = 1/2[(n/2)th term + {(n/2) + 1}th] term when n is even
Range = Maximum value – Minimum value
Calculation:
Arranging the given data in ascending order
2, 3, 3, 4, 4, 4, 5, 6, 8, 9, 9, 10, 11, 15, 19
Here, Most frequent data is 4 so
Mode = 4
Total terms in the given data, (n) = 15 (It is odd)
Median = {(n + 1)/2}th term when n is odd
⇒ {(15 + 1)/2}th term
⇒ (8)th term
⇒ 6
Now, Range = Maximum value – Minimum value
⇒ 19 – 2 = 17
Mean of Range, Mode and median = (Range + Mode + Median)/3
⇒ (17 + 4 + 6)/3
⇒ 27/3 = 9
∴ The mean of the Range, Mode and Median is 9
Find the mean of given data:
class interval | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |
Frequency | 9 | 13 | 6 | 4 | 6 | 2 | 3 |
Answer (Detailed Solution Below)
Statistics Question 7 Detailed Solution
Download Solution PDFFormula used:
The mean of grouped data is given by,
\(\bar X\ = \frac{∑ f_iX_i}{∑ f_i}\)
Where, \(u_i \ = \ \frac{X_i\ -\ a}{h}\)
Xi = mean of ith class
fi = frequency corresponding to ith class
Given:
class interval | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |
Frequency | 9 | 13 | 6 | 4 | 6 | 2 | 3 |
Calculation:
Now, to calculate the mean of data will have to find ∑fiXi and ∑fi as below,
Class Interval | fi | Xi | fiXi |
10 - 20 | 9 | 15 | 135 |
20 - 30 | 13 | 25 | 325 |
30 - 40 | 6 | 35 | 210 |
40 - 50 | 4 | 45 | 180 |
50 - 60 | 6 | 55 | 330 |
60 - 70 | 2 | 65 | 130 |
70 - 80 | 3 | 75 | 225 |
∑fi = 43 | ∑Xi = 315 | ∑fiXi = 1535 |
Then,
We know that, mean of grouped data is given by
\(\bar X\ = \frac{∑ f_iX_i}{∑ f_i}\)
= \(\frac{1535}{43}\)
= 35.7
Hence, the mean of the grouped data is 35.7
If mean and mode of some data are 4 & 10 respectively, its median will be:
Answer (Detailed Solution Below)
Statistics Question 8 Detailed Solution
Download Solution PDFConcept:
Mean: The mean or average of a data set is found by adding all numbers in the data set and then dividing by the number of values in the set.
Mode: The mode is the value that appears most frequently in a data set.
Median: The median is a numeric value that separates the higher half of a set from the lower half.
Relation b/w mean, mode and median:
Mode = 3(Median) - 2(Mean)
Calculation:
Given that,
mean of data = 4 and mode of data = 10
We know that
Mode = 3(Median) - 2(Mean)
⇒ 10 = 3(median) - 2(4)
⇒ 3(median) = 18
⇒ median = 6
Hence, the median of data will be 6.
If the standard deviation of 0, 1, 2, 3 ______ 9 is K, then the standard deviation of 10, 11, 12, 13 _____ 19 will be:
Answer (Detailed Solution Below)
Statistics Question 9 Detailed Solution
Download Solution PDFFormula Used∶
- σ2 = ∑(xi – x)2/n
- Standard deviation is same when each element is increased by the same constant
Calculation:
Since each data increases by 10,
There will be no change in standard deviation because (xi – x) remains same.
∴ The standard deviation of 10, 11, 12, 13 _____ 19 will be will be K.
Alternate Method
Find the median of the given set of numbers 2, 6, 6, 8, 4, 2, 7, 9
Answer (Detailed Solution Below)
Statistics Question 10 Detailed Solution
Download Solution PDFConcept:
Median: The median is the middle number in a sorted- ascending or descending list of numbers.
Case 1: If the number of observations (n) is even
\({\rm{Median\;}} = {\rm{\;}}\frac{{{\rm{value\;of\;}}{{\left( {\frac{{\rm{n}}}{2}} \right)}^{{\rm{th}}}}{\rm{\;observation\;}} + {\rm{\;\;value\;of\;}}{{\left( {\frac{{\rm{n}}}{2}{\rm{\;}} + 1} \right)}^{{\rm{th}}}}{\rm{\;observation}}}}{2}\)
Case 2: If the number of observations (n) is odd
\({\rm{Median\;}} = {\rm{value\;of\;}}{\left( {\frac{{{\rm{n}} + 1}}{2}} \right)^{{\rm{th}}}}{\rm{\;observation}}\)
Calculation:
Given values 2, 6, 6, 8, 4, 2, 7, 9
Arrange the observations in ascending order:
2, 2, 4, 6, 6, 7, 8, 9
Here, n = 8 = even
As we know, If n is even then,
\({\rm{Median\;}} = {\rm{\;}}\frac{{{\rm{value\;of\;}}{{\left( {\frac{{\rm{n}}}{2}} \right)}^{{\rm{th}}}}{\rm{\;observation\;}} + {\rm{\;\;value\;of\;}}{{\left( {\frac{{\rm{n}}}{2}{\rm{\;}} + 1} \right)}^{{\rm{th}}}}{\rm{\;observation}}}}{2}\)
= \(\rm \frac{4^{th} \;\text{observation}+5^{th} \;\text{observation}}{2} \)
= \(\frac{6+6}{2} =6\)
Hence Median = 6
What is the standard deviation of the observations
\(-\sqrt{6}, -\sqrt{5},- \sqrt{4}, -1, 1, \sqrt{4}, \sqrt{5}, \sqrt{6} \ ?\)
Answer (Detailed Solution Below)
Statistics Question 11 Detailed Solution
Download Solution PDFConcept:
Standard deviation:
The standard deviation of the observation set \(\rm \{x_i,i=1,2,3,\cdots\}\) is given as follows:
\(\rm \sigma=\sqrt{\dfrac{\sum\left(x_i-\mu\right)^2}{N}}\)
Where \(\rm N=\mbox{size of the observation set}\) and \(\rm \mu=\mbox{mean of the observations}\) .
Calculations:
First, we will calculate the mean of the given observations.
\(\begin{align*} \mu &= \dfrac{-\sqrt6-\sqrt5-\sqrt4-1+1+\sqrt4+\sqrt5+\sqrt6}{8}= 0 \end{align*}\)
Therefore, the numerator inside the square root term of the standard deviation formula will simply be equal to \(\rm (x_i-\mu)^2=x_i^2\) .
Now we observe that \(\rm N=8\) .
Therefore, the standard deviation is given as follows:
\(\begin{align*} \sigma &= \sqrt{\dfrac{\left(-\sqrt6\right)^2+\left(-\sqrt5\right)^2+\left(-\sqrt4\right)^2+\left(-1\right)^2+\left(1\right)^2+\left(\sqrt4\right)^2+\left(\sqrt5\right)^2+\left(\sqrt6\right)^2}{8}}\\ &= \sqrt{\dfrac{32}{8}}\\ &= \sqrt4\\ &= 2 \end{align*}\)
Therefore, the standard deviation of the given observations is 2.
The mean of four numbers is 37. The mean of the smallest three of them is 34. If the range of the data is 15, what is the mean of the largest three?
Answer (Detailed Solution Below)
Statistics Question 12 Detailed Solution
Download Solution PDFCalculation:
Let the numbers be x1, x2, x3, x4.
The mean of four numbers x1, x2, x3, x4 = 37
The sum of four numbers x1, x2, x3, x4 = 37 × 4 = 148.
The mean of the smallest three numbers x1, x2, x3 = 34
The sum of the smallest three numbers x1, x2, x3 = 34 × 3 = 102.
∴ The value of the largest number x4 = 148 – 102 = 46.
The Range (Difference between largest and smallest value) x4 – x1 = 15.
∴ Smallest number x1 = 46 – 15 = 31.
Now,
The sum of x2, x3 = Total sum – (sum of smallest and largest number).
⇒ 148 – (46 + 31)
⇒ 148 – 77
⇒ 71
Now,
The mean of the Largest three numbers x2, x3, x4 = (71 + 46)/3 = 117/3 = 39If the mean of a frequency distribution is 100 and the coefficient of variation is 45%, then what is the value of the variance?
Answer (Detailed Solution Below)
Statistics Question 13 Detailed Solution
Download Solution PDFConcept:
Coefficient of variation = \(\rm\text{Standard Deviation} \over\text{ Mean}\)
Variance = (Standard Deviation)2
Calculation:
Given coefficient of variation = 45% = 0.45
And mean = 100
As Coefficient of variation = \(\rm\text{Standard Deviation} \over\text{ Mean}\)
0.45 = \(\rm\text{Standard Deviation} \over100\)
Standard Deviation = 100 × 0.45
SD = 45
∴ Variance = 452 = 2025
The data given below shows the marks obtained by various students.
Marks |
Number of students |
10 – 12 |
6 |
12 – 14 |
8 |
14 – 16 |
5 |
16 – 18 |
7 |
18 - 20 |
4 |
What is the mean marks (Correct up to two decimal places) of given data?
Answer (Detailed Solution Below)
Statistics Question 14 Detailed Solution
Download Solution PDF\(\bar x\left( {mean} \right)\; = \;\frac{{\sum fx}}{n}\)
⇒ n = total frequency
\(\sum fx = Sum\;of\;the\;product\;of\;mid - interval\;values\;and\;their\;corresponding\;frequency\;\;\)
Mid value of 10 – 12 = (10 + 12)/2 = 11
Mid value of 12 – 14 = (12 + 14 )/2 = 13
Mid value of 14 – 16 = (14 + 16 )/2 = 15
Mid value of 16 – 18 = (16 + 18 )/2 = 17
Mid value of 18 – 20 = (18 + 20 )/2 = 19
\(\Rightarrow \;Mean\; = \;\frac{{11\; \times \;6\; + \;13\; \times \;8\; + \;15\; \times \;5\; + \;17\; \times \;7\; + \;19\; \times \;4}}{{6\; + \;8\; + \;5\; + \;7\; + \;4}} = \;\frac{{440}}{{30}}\)
⇒ Mean = 14.67
∴The mean marks of the given data are 14.67
If the mode of the following data is 7, then the value of k in the data set 3, 8, 6, 7, 1, 6, 10, 6, 7, 2k + 5, 9, 7, and 13 is:
Answer (Detailed Solution Below)
Statistics Question 15 Detailed Solution
Download Solution PDFConcept:
Mode is the value that occurs most often in the data set of values.
Calculation:
Given data values are 3, 8, 6, 7, 1, 6, 10, 6, 7, 2k + 5, 9, 7, and 13
In the above data set, values 6, and 7 have occurred more times i.e., 3 times
But given that mode is 7.
So, 7 should occur more times than 6.
Hence the variable 2k + 5 must be 7
⇒ 2k + 5 = 7
⇒ 2k = 2
∴ k = 1