Mathematics · Quantitative Aptitude
Statistics and Dispersion
515 Questions
Statistics and dispersion involve the calculation of mean, standard deviation, variance, and coefficient of variation for data sets. These questions also cover probability distributions and cumulative frequency analysis. Such quantitative aptitude topics are heavily featured in banking and SSC examinations.
Standard deviationNormal distributionMean calculationCumulative frequencyCoefficient of variation
Statistics and Dispersion Questions
The mean of a data set consisting of $20$ observations is $40$. If one observation $53$ was wrongly recorded as $33$, then the correct mean will be:
A
Correct answer
Explanation
$Mean=\dfrac{S}{n}$
where S=sum of all observations
n=number of observations
hence, $40=\dfrac{S}{20}\Rightarrow S=800$
but Since, 53 is recorded as 33 so we need to add $(53-33=20)$ to get the correct mean which is
$Mean _{correct}=\dfrac{800+20}{20}=41$
Which of the following is NOT a property of the variance?
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It is always positive
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It is a constant
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It is always negative
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It is equal to the square of the standard deviation
C
Correct answer
Explanation
The variance is not always negative. It can be positive or zero, depending on the distribution of the random variable.
If a random variable X follows a normal distribution with a mean of 50 and a standard deviation of 10, what is the probability that X will be greater than 60?
A
Correct answer
Explanation
To find the probability that X will be greater than 60, we first standardize X using the formula Z = (X - μ) / σ, where μ is the mean and σ is the standard deviation. In this case, Z = (60 - 50) / 10 = 1. Then, we look up the probability that a standard normal random variable will be greater than 1 in the standard normal distribution table or use a calculator. This probability is approximately 0.1587.
The mean and standard deviation of a normal distribution are 40 and 5, respectively. What is the probability that a randomly selected value from this distribution will be between 30 and 50?
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0.3413
-
0.6826
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0.8413
-
0.9545
B
Correct answer
Explanation
To find the probability that a randomly selected value from this distribution will be between 30 and 50, we first standardize the values using the formula Z = (X - μ) / σ, where μ is the mean and σ is the standard deviation. In this case, Z1 = (30 - 40) / 5 = -2 and Z2 = (50 - 40) / 5 = 2. Then, we look up the probability that a standard normal random variable will be between -2 and 2 in the standard normal distribution table or use a calculator. This probability is approximately 0.6826.
If X is a random variable following a normal distribution with a mean of 50 and a standard deviation of 10, what is the probability that X will be between 40 and 60?
-
0.3413
-
0.6826
-
0.8413
-
0.9545
B
Correct answer
Explanation
To find the probability that X will be between 40 and 60, we first standardize the values using the formula Z = (X - μ) / σ, where μ is the mean and σ is the standard deviation. In this case, Z1 = (40 - 50) / 10 = -1 and Z2 = (60 - 50) / 10 = 1. Then, we look up the probability that a standard normal random variable will be between -1 and 1 in the standard normal distribution table or use a calculator. This probability is approximately 0.6826.
If X is a random variable following a normal distribution with a mean of 50 and a standard deviation of 10, what is the probability that X will be less than 40?
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0.0228
-
0.05
-
0.1587
-
0.3413
A
Correct answer
Explanation
To find the probability that X will be less than 40, we first standardize the value using the formula Z = (X - μ) / σ, where μ is the mean and σ is the standard deviation. In this case, Z = (40 - 50) / 10 = -1. Then, we look up the probability that a standard normal random variable will be less than -1 in the standard normal distribution table or use a calculator. This probability is approximately 0.0228.
A company claims that the weights of its products are normally distributed with a mean of 100 grams and a standard deviation of 10 grams. If a random sample of 100 products is selected, what is the probability that the average weight of the sample will be less than 95 grams?
-
0.0228
-
0.05
-
0.1587
-
0.3413
A
Correct answer
Explanation
By the Central Limit Theorem, the distribution of sample means will be approximately normal with a mean equal to the population mean and a standard deviation equal to the population standard deviation divided by the square root of the sample size. In this case, the mean of the sample means will be 100 grams and the standard deviation of the sample means will be 10 grams / √100 = 1 gram. To find the probability that the average weight of the sample will be less than 95 grams, we standardize the value using the formula Z = (X - μ) / σ, where μ is the mean and σ is the standard deviation. In this case, Z = (95 - 100) / 1 = -5. Then, we look up the probability that a standard normal random variable will be less than -5 in the standard normal distribution table or use a calculator. This probability is approximately 0.0228.
A random variable X follows a normal distribution with mean μ = 10 and standard deviation σ = 2. What is the probability that X is between 6 and 14?
-
0.6827
-
0.9545
-
0.3173
-
0.0455
A
Correct answer
Explanation
The probability that X is between 6 and 14 is given by the area under the normal distribution curve between these values. We can use the standard normal distribution (Z) to calculate this probability: P(6 ≤ X ≤ 14) = P((6 - 10)/2 ≤ Z ≤ (14 - 10)/2) = P(-2 ≤ Z ≤ 2) = 0.6827.
A continuous random variable X has a probability density function given by f(x) = 3x^2 for 0 ≤ x ≤ 1. Find the variance of X.
A
Correct answer
Explanation
The variance of X is given by Var(X) = E(X^2) - [E(X)]^2. First, we find E(X^2): E(X^2) = ∫[0, 1] 3x^2 * x dx = ∫[0, 1] 3x^3 dx = [3x^4/4]_[0, 1] = 3/4. Then, we find [E(X)]^2: [E(X)]^2 = (3.5)^2 = 12.25. Therefore, Var(X) = 3/4 - 12.25 = 0.25.
In ANOVA, the degrees of freedom for the between-groups variance is:
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k - 1
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n - k
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k(n - 1)
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n(k - 1)
A
Correct answer
Explanation
The degrees of freedom for the between-groups variance is k - 1, where k is the number of groups.
What is the variance of the Poisson distribution?
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lambda
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lambda^2
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lambda^3
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lambda^4
A
Correct answer
Explanation
The variance of the Poisson distribution is also equal to the average rate of occurrence, which is denoted by lambda.
What is the formula for calculating the test-retest reliability coefficient?
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$$r_{tt} = \frac{\sum(X_1 - \bar{X})(X_2 - \bar{X})}{\sqrt{\sum(X_1 - \bar{X})^2\sum(X_2 - \bar{X})^2}}$$
-
$$r_{tt} = \frac{\sum(X_1 - \bar{X})(X_2 - \bar{X})}{\sqrt{\sum(X_1 - \bar{X})^2\sum(X_2 - \bar{X})^2}}$$
-
$$r_{tt} = \frac{\sum(X_1 - \bar{X})(X_2 - \bar{X})}{\sqrt{\sum(X_1 - \bar{X})^2\sum(X_2 - \bar{X})^2}}$$
-
$$r_{tt} = \frac{\sum(X_1 - \bar{X})(X_2 - \bar{X})}{\sqrt{\sum(X_1 - \bar{X})^2\sum(X_2 - \bar{X})^2}}$$
A,B,C,D
Correct answer
Explanation
The formula for calculating the test-retest reliability coefficient is $$r_{tt} = \frac{\sum(X_1 - \bar{X})(X_2 - \bar{X})}{\sqrt{\sum(X_1 - \bar{X})^2\sum(X_2 - \bar{X})^2}}$$, where X_1 and X_2 are the scores of the same group of test takers on two different administrations of the test, and (\bar{X}) is the mean score on the test.
What is the formula for calculating the variance of a random variable representing claim severity?
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$$V(X) = E(X^2) - (E(X))^2$$
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$$V(X) = E(X) - (E(X^2))^2$$
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$$V(X) = (E(X))^2 - E(X^2)$$
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$$V(X) = E(X^2) + (E(X))^2$$
A
Correct answer
Explanation
The variance of a random variable is calculated as the expected value of the squared deviation from the mean.
What is the average HDI of the world?
A
Correct answer
Explanation
The average HDI of the world is approximately 0.7, which is considered to be a medium level of human development.
What is the mean of a normal distribution with a mean of 5 and a standard deviation of 2?
A
Correct answer
Explanation
The mean of a normal distribution is the center of the distribution. In this case, the mean is 5.