[Solved] Normal Distribution MCQ [Free PDF] - Objective Question Answer for Normal Distribution Quiz - Download Now! (2024)

1. ∑(X_i) for i = 1 to n
2. ∑(X_i - X̄)² for i = 1 to n, where X̄ is the sample mean
3. min(X_i) for i = 1 to n
4. max(X_i) for i = 1 to n

Explanation -

In the case of the normal distribution, a sufficient statistic for μ (the mean of the distribution) is the sum of the sample observations (∑X_ifor i = 1 to n), because the mean value directly depends on the sum of all the observations.

On the other hand, the minimum value of the sample, min(X_i) for i = 1 to n, and the maximum value of the sample, max(X_i) for i = 1 to n, do not hold all necessary information to calculate μ, and thus they are not sufficient statistics.

In this question, the minimum value (min(X_i)) is listed as the statistic that is not sufficient for calculating μ.

Hence option (iii) is correct.

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1. Mean = Variance = 1
2. Standard deviation = Variance = 0
3. Mean = 0, Variance = 1
4. Mean = Standard deviation
5. Not Attempted

Explanation:

Normal/Gaussian/Bell distribution:

Probability distribution function (PDF) for a normal distribution is:

\(PDF = f\left( x \right) = \frac{1}{{\sqrt {2\pi {\sigma ^2}\;} }}{e^{ - \frac{1}{2}{{\left( {\frac{{x - \mu }}{\sigma }} \right)}^2}}}\)

Where,

x = normal random variable

μ = mean = mode = median

σ = standard deviation. σ2= variance

For continuous distribution

Standard normal distribution:

In this case,\(\frac{{x\; - \;\mu }}{\sigma } = z\)

where z is a standard normal variable

Mean = Mode = Median = 0

Standard deviation = 1

Variance = 1

Explanation:

From the above discussion, we can say that,

Mean = 0 and Variance = 1

Hence, option 3 is correct.

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1. standard variation
2. mean
3. covariance
4. variance

The correct answer is (option 2) Mean.

Explanation:

Normal Distribution:

• Normal distribution is also known as the Gaussian distribution.
• It is aprobability distributionthat is symmetric about the mean,
  showing that data near the mean are more frequent in occurrence than data far from the mean.
• In graphical form, the normal distribution appears as a "bell curve".

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1. bell shaped
2. triangular
3. rectangular
4. circular

Correct option 1

Concept:

Normal curve:

The Normal Distribution is defined by theprobability density functionfor a continuous random variable in a system. Let us say, f(x) is the probability density function and X is the random variable. Hence, it defines a function that is integrated between the range or interval (x to x + dx), giving the probability of random variable X, by considering the values between x and x+dx.

f(x) ≥ 0 ∀ x ϵ (−∞,+∞)

And-∞∫+∞f(x) = 1

In probability theory and statistics, theNormal Distribution, also called theGaussian Distribution, is the most significant continuous probability distribution. Sometimes it is also called a bell curve. A large number of random variables are either nearly or exactly represented by the normal distribution.

the shape of a Normal distribution curve or bell curve

Normal distributions have the following features

• Symmetric bell shape
• Mean and median are equal; both are located at the center of the distribution
• $\approx 68\mathrm{\%}$≈68%approximately equals, 68, percentof the data falls within$1$11the standard deviation of the mean
• $\approx 95\mathrm{\%}$≈95%approximately equals, 95, percentof the data falls within$2$22standard deviations of the mean
• $\approx 99.7\mathrm{\%}$≈99.7%approximately equals, 99, points, 7, percentof the data falls within$3$33standard deviations of the mean

Additional point

• 1. The shape of the distribution is determined by the average, μ (orX), and the standard deviation, σ.
• The highest point on the curve is the average.
• The distribution is symmetrical about the average.
• As you move away from the average, the points occur with less frequency.
• Most of the area under the curve (99.7%) lies between -3σ and +3σ of the average.

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1. Mean = Variance = 1
2. Standard deviation = Variance = 0
3. Mean = 0, Variance = 1
4. Mean = Standard deviation

Explanation:

Normal/Gaussian/Bell distribution:

Probability distribution function (PDF) for a normal distribution is:

\(PDF = f\left( x \right) = \frac{1}{{\sqrt {2\pi {\sigma ^2}\;} }}{e^{ - \frac{1}{2}{{\left( {\frac{{x - \mu }}{\sigma }} \right)}^2}}}\)

Where,

x = normal random variable

μ = mean = mode = median

σ = standard deviation. σ2= variance

For continuous distribution

Standard normal distribution:

In this case,\(\frac{{x\; - \;\mu }}{\sigma } = z\)

where z is a standard normal variable

Mean = Mode = Median = 0

Standard deviation = 1

Variance = 1

Explanation:

From the above discussion, we can say that,

Mean = 0 and Variance = 1

Hence, option 3 is correct.

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Normal Distribution Question 6

1. 0
2. \(\frac{1}{2}\)
3. \(1\;-\;\frac{1}{e}\)
4. 1

Normal Distribution Question 6 Detailed Solution

Concept:

Normal/Gaussian/Bell distribution:

Probability distribution function (PDF) for a normal distribution is:

\(PDF = f\left( x \right) = \frac{1}{{\sqrt {2\pi {σ ^2}\;} }}{e^{ - \frac{1}{2}{{\left( {\frac{{x - μ }}{σ }} \right)}^2}}}\)

where

x = normal random variable

μ = mean = mode = median

σ = standard deviation and σ²= variance.

Note:

1) Normal distribution is symmetric about its mean.

Calculation:

Given:

\(f(x)=\frac{1}{\sqrt{8\pi}}e^{-\left \{ \frac{(x\;-\;1)^2}{8} \right \}}\)

Comparing it with the standard normal distribution

μ = mean = 1 andσ² = variance = 4

The distribution function is divided into two equal parts which are equiprobable.

\(\therefore\; \int_{-\infty}^{\infty}f(x)dx=1\)

\(\therefore\; \int_{-\infty}^{1}f(x)dx\;+\;\int_{1}^{\infty}f(x)dx=1\)

∵ A normal distribution is symmetric about mean i.e.\(\therefore\; \int_{-\infty}^{1}f(x)dx\;=\;\int_{1}^{\infty}f(x)dx\)

\(\therefore\; \;\int_{1}^{\infty}f(x)dx=\frac{1}{2}\)

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Normal Distribution Question 7

1. I
2. II
3. III
4. IV

Normal Distribution Question 7 Detailed Solution

Probability distribution function for normal distribution is given by

\(f\left( x \right) = \frac{1}{{\sigma \surd \pi }}{e^ - }\frac{{{{\left( {x - \mu } \right)}^2}}}{{2{\sigma ^2}}}\)

Variance = σ²

For lowest variance, σ should be lowest

And if σ decreased, f (x) will increase

So, curve will have highest peak.

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Normal Distribution Question 8

1. < 50%
2. 50%
3. 75%
4. 100 %

Normal Distribution Question 8 Detailed Solution

Concept:

• The normal distribution is a probability function that describes how the values of a variable are distributed.
• For a normally distributed variable xwith mean μ and standard deviationσ, the normal variate z is given by the formula:\(\rm z = \dfrac{x - \mu}{\sigma}\).

Calculation:

Given

Standard deviation σ = 200 mm, Mean μ = 1000 mm.

For x = 1200,\(\rm z = \dfrac{x - \mu}{\sigma}=\dfrac{1200-1000}{200}=1\)

P (X > 1200 mm) = P ( z > 1)

z is normal variate,

We know (P ( - 1 < Z < 1 ) = 0.68 (i.e. 68% of data is within one standard deviation of mean)

P (0 < Z < 1 ) =0.68/2 = 0.34

So P (z >1) = 0.5- 0.34 = 0.16 % < 50 %

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Normal Distribution Question 9

1. standard variation
2. mean
3. covariance
4. variance

Normal Distribution Question 9 Detailed Solution

The correct answer is (option 2) Mean.

Explanation:

Normal Distribution:

• Normal distribution is also known as the Gaussian distribution.
• It is aprobability distributionthat is symmetric about the mean,
  showing that data near the mean are more frequent in occurrence than data far from the mean.
• In graphical form, the normal distribution appears as a "bell curve".

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Normal Distribution Question 10

1. a
2. 2a
3. 1 - 2a
4. 1 - a

Normal Distribution Question 10 Detailed Solution

Concept:

For |X|≤ a

-a≤ X≤ a

sum of all event's probability = 1

Also, for normal distribution

Z-value (Z) =\(\frac {X-μ }{σ }\)

Calculation:

Given:

μ = 0; variance (v) = 9;P(X≥ 3) = a;

σ =√v =√9

σ= 3

Z =\(\frac {X-μ }{σ }\)=\(\frac {3\ -\ 0 }{3 }\)

Z = 1

Hence, a normal distribution with Z = 1

⇒P(X≥ 3) = P(X ≤ 3) = a

Max. probability of any event = 1

P(-∞ ≤ X ≤ ∞) = 1

P(-∞ ≤ X ≤ ∞) =P(X≤ 3) + P(-3 ≤ X ≤3) + P(X ≥ 3)

1 =P(X≤ 3) + P(|X|≤3) + P(X ≥ 3)

1 = a +P(|X|≤3) + a

P(|X|≤3) = 1 - 2a

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Normal Distribution Question 11

Normal Distribution Question 11 Detailed Solution

Concept:

The normal distribution is symmetric with respect to the mean position and the standard normal variate is given by the relation.

Let variate be x

\({\rm{z}} = \frac{{{\rm{x}} - {\rm{M}}}}{{\rm{σ }}}\)

Where z =Normal variate, M =Mean,σ =Standard deviation

For P(z ≥ 0) = P(z < 0) = 0.5 or 50%

Calculation:

Given:

M = 500

σ = 50

P(x > 500) = ?

Where X follows the normal distribution

We know that standard normal variable

\({\rm{z}} = \frac{{{\rm{x}} - {\rm{M}}}}{{\rm{σ }}}\)

converting into normalized form

x = 50z + 500

given, x > 500, therefore

50z + 500 > 500 or z > 0

probability, P(z > 0) = 0.5

∴P(x >500) = 0.5 or 50%

For x = 500, the standard normal variate is

\({\rm{z}} = \frac{{500 - 500}}{{50}} = 0\)

∴ P(x> 500) = P(z > 0) = 0.50 or 50%(See figure)

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Normal Distribution Question 12

Normal Distribution Question 12 Detailed Solution

It is given that mean = 0 and variance = σ². It means it represents the standard normal distribution and in case of standard normal distribution graph is symmetric. In symmetric case, mean = median = mode. As, it is given that mean = 0. So, median will also be 0.

Alternate Solution

Median is a point where the probability of getting less than median is ½ and probability of getting greater than median is ½.

Consider a continuous random variable X with median 0 which means

\(P\left( {X < = 0} \right) = \frac{1}{2}\;\;and\;P\left( {X > 0} \right) = \frac{1}{2}\)

Here variable Y can take only non-negative values because Y = max(X, 0).

So, since (X <= 0) → (Y = 0) and (X > 0) → (Y > 0), it means P (Y = 0) = ½ and P(Y > 0) = ½.

It matches with the definition of median of a random variable.

Therefore, median of random variable Y is 0.

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Normal Distribution Question 13

1. bell shaped
2. triangular
3. rectangular
4. circular

Normal Distribution Question 13 Detailed Solution

Correct option 1

Concept:

Normal curve:

The Normal Distribution is defined by theprobability density functionfor a continuous random variable in a system. Let us say, f(x) is the probability density function and X is the random variable. Hence, it defines a function that is integrated between the range or interval (x to x + dx), giving the probability of random variable X, by considering the values between x and x+dx.

f(x) ≥ 0 ∀ x ϵ (−∞,+∞)

And-∞∫+∞f(x) = 1

In probability theory and statistics, theNormal Distribution, also called theGaussian Distribution, is the most significant continuous probability distribution. Sometimes it is also called a bell curve. A large number of random variables are either nearly or exactly represented by the normal distribution.

the shape of a Normal distribution curve or bell curve

Normal distributions have the following features

• Symmetric bell shape
• Mean and median are equal; both are located at the center of the distribution
• $\approx 68\mathrm{\%}$≈68%approximately equals, 68, percentof the data falls within$1$11the standard deviation of the mean
• $\approx 95\mathrm{\%}$≈95%approximately equals, 95, percentof the data falls within$2$22standard deviations of the mean
• $\approx 99.7\mathrm{\%}$≈99.7%approximately equals, 99, points, 7, percentof the data falls within$3$33standard deviations of the mean

Additional point

• 1. The shape of the distribution is determined by the average, μ (orX), and the standard deviation, σ.
• The highest point on the curve is the average.
• The distribution is symmetrical about the average.
• As you move away from the average, the points occur with less frequency.
• Most of the area under the curve (99.7%) lies between -3σ and +3σ of the average.

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Normal Distribution Question 14

1. nσ
2. σ
3. \(\sigma \over n\)
4. \(\sigma \over \sqrt{n} \)

Normal Distribution Question 14 Detailed Solution

Explanation:

Sampling Distribution of Sampling Mean (\(\bar{X}\))

If repeated random samples of a given sizenare taken from a population of values for a quantitative variable, where the population mean isμand the population standard deviation is σ.

• The meanof all sample means (\(\bar{X}\))is the population mean μ.

As for the spread of all sample means, theCentral Limit theoremdictates the behaviour much more precisely than saying that there is less spread for larger samples. In fact, the standard deviation of all sample means is directly related to the sample size,nas indicated below.

• The standard deviation of all sample means (\(\bar{X}\)) is\(\mathbf{\sigma \over\sqrt{n}}\)

Note: the square root of sample sizenappears in the denominator, the standard deviation does decrease as the sample size increases.

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1. \(\frac{4}{5}\)
2. \(\frac{2}{5}\)
3. \(\frac{9}{10}\)
4. \(\frac{3}{5}\)

Concept:

\(\rm mean = \frac{{1 + 2 + 3 + \ldots \cdots \cdots + 10}}{{10}} = 5.5\)

Given, P(X ≤4) =\(\frac{{1}}{{5}}\), From symmetricity ofdistribution, P(X≥ 7) =\(\frac{{1}}{{5}}\)

we know that total probability is equal to one hence

\(\rm P(-\infty < x ≤ 4) + P(4 < x < 7) + P(7 ≤ x ≤ \infty) = 1\)

\(\rm \begin{array}{l} \Rightarrow \frac{1}{5} + P\left( {4 < x < 7} \right) + \frac{1}{5} = 1\\ \rm P\left( {4 < x < 7} \right) = 1 - \frac{2}{5} = \frac{3}{5} \end{array}\)

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