Mean and the first quartile for a normal distribution are 11 and 3 respectively. Which of the following is the

Question 15 of 46

11

(a)

8

(b)

14

(c)

19

(d)

10

Gujarat GsebGujarat Board (GSEB) HSC Commerce Board 2020MCQ· 1mImportance★★★★★

33% · 15/46 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Symmetry gives $Q_3 = 2\mu - Q_1 = 22 - 3 = 19$ — option (c).

For a normal distribution the mean equals the median, and $Q_1$ and $Q_3$ are placed symmetrically about the mean at equal distances:

$\mu - Q_1 = Q_3 - \mu$

So the mean is the midpoint of $Q_1$ and $Q_3$ :

$\mu = \frac{Q_1 + Q_3}{2} \implies Q_3 = 2\mu - Q_1$

Substituting $\mu = 11$ , $Q_1 = 3$ : …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 15 on this concept.

CBSE 2026Set MARCH1 markMCQ
$Q. In normal distribution, usually which limits include 99% of the observations?$
(a) $\mu \pm 1.96\sigma$
(b) $\mu \pm 2\sigma$
(c) $\mu \pm 3\sigma$
(d) $\mu \pm 2.57\sigma$
›Reveal solutionSolution

About $99\%$ of a normal distribution lies within $\mu\pm2.57\sigma$ (about $2.58\sigma$ ).

Standard normal area results: $\mu\pm1.96\sigma$ covers about $95\%$ , $\mu\pm3\sigma$ covers about $99.73\%$ , and $\mu\pm2.57\sigma$ (more precisely $2.58\sigma$ ) covers about …

Start your 14-day free trial to see the full solution →
CBSE 2026Set MARCH1 markMCQ
$Q. For a normal distribution, approximate value of mean deviation is 20. Which of the following is the value of quartile deviation?$
(a) $\dfrac{25}{3}$
(b) $\dfrac{32}{3}$
(c) $24$
(d) $\dfrac{50}{3}$
›Reveal solutionSolution

Using $QD:MD:SD=10:12:15$ , $QD=\tfrac{10}{12}(20)=\tfrac{50}{3}$ .

For a normal distribution the approximate relations are $QD\approx\tfrac23\sigma$ and $MD\approx\tfrac45\sigma$ , giving $QD:MD:SD=10:12:15$ . Therefore

$QD=\frac{10}{12}\times MD=\frac{10}{12}\times20=\frac{200}{12}=\frac{50}{3}.$ …

Start your 14-day free trial to see the full solution →
CBSE 2026Set MARCH1 mark
$Q. What is the shape of normal curve?$
›Reveal solutionSolution

The normal curve is a symmetric, single-peaked, bell-shaped curve.

The graph of the normal distribution is a smooth, bell-shaped curve. It is perfectly symmetric about its mean $\mu$ (where mean $=$ median $=$ mode), rises to a single maximum at $x=\mu$ , and falls off equally on both sides. Its two tails extend to $-\infty$ and $+\infty$ …

Start your 14-day free trial to see the full solution →
CBSE 2025Set MARCH1 markMCQ
$X = \mu$
(a) $0$
(b) $0.5$
(c) $1$
(d) $-0.5$
›Reveal solutionSolution

The normal curve is symmetric about $X = \mu$ , so the area on either side of the mean is $0.5$ — option (b).

GSEB Class-12 Statistics, Normal Distribution:

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set MARCH1 markMCQ
$Q. Mean and the first quartile for a normal distribution are 15 and 11 respectively. Which of the following is the value of the third quartile?$
(a) $8$
(b) $14$
(c) $19$
(d) $10$
›Reveal solutionSolution

By the symmetry of the normal curve, $Q_3 = 2\mu - Q_1 = 2(15) - 11 = 19$ — option (c).

GSEB Class-12 Statistics, Normal Distribution:

For a normal distribution the quartiles are placed symmetrically about the mean:

$Q_1 = \mu - 0.6745\,\sigma, \qquad Q_3 = \mu + 0.6745\,\sigma$

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set MARCH1 mark
$Q. The approximate value of mean deviation for a normal distribution is 12. Find its standard deviation.$
›Reveal solutionSolution

With M.D. $\approx \tfrac{4}{5}\sigma = 12$ , the standard deviation is $\sigma = \tfrac{5}{4}\times 12 = 15$ .

GSEB Class-12 Statistics, Normal Distribution:

A property of the normal distribution is that its mean deviation is approximately:

$\text{M.D.} = \frac{4}{5}\,\sigma$

…

Start your 14-day free trial to see the full solution →
CBSE 2023Set MARCH1 markMCQ
$X = \mu$
(a) (A) $0$
(b) (B) $-0.5$
(c) (C) $1$
(d) (D) $0.5$
›Reveal solutionSolution

By symmetry of the normal curve about the mean, the area to the left of $X=\mu$ is half of the total, i.e. $0.5$ . Option (D).

The total area under a normal curve equals $1$ . The curve is perfectly symmetric about the mean $\mu$ , so the vertical line at $X=\mu$ divides the total area i …

Start your 14-day free trial to see the full solution →
CBSE 2023Set MARCH1 markMCQ
$Q. Which of the following is the approximate value of mean deviation for normal variable?$
(a) (A) $\frac{4}{5}\sigma$
(b) (B) $\frac{4}{5}\mu$
(c) (C) $\frac{2}{3}\sigma$
(d) (D) $\frac{2}{3}\mu$
›Reveal solutionSolution

The mean deviation of a normal variable is approximately $\dfrac{4}{5}\sigma$ . Option (A).

Among the standard measures of dispersion for a normal distribution, the mean deviation (about the mean) is a fixed multiple of the standard deviation. Its approximate value is

$\text{M.D.}\approx\frac{4}{5}\sigma\;(=0.8\,\sigma),$ …

Start your 14-day free trial to see the full solution →
CBSE 2023Set MARCH1 mark
$Q. What is the shape of normal curve?$
›Reveal solutionSolution

The normal curve is a symmetric, bell-shaped, unimodal curve.

The graph of a normal distribution is a smooth, continuous, bell-shaped curve. It is symmetric about its mean $\mu$ , has a single peak (unimodal) at $x=\mu$ , and its two tails extend indefinitely on both sides, …

Start your 14-day free trial to see the full solution →
CBSE 2022Set MARCH1 markMCQ
$\mu$
(a) 0
(b) 0.5
(c) 1
(d) $-0.5$
›Reveal solutionSolution

Because the normal curve is symmetric about $\mu$ , the area to the left of the perpendicular at $\mu$ is exactly half the total area, i.e. $\mathbf{0.5}$ .

…

Start your 14-day free trial to see the full solution →
CBSE 2022Set MARCH1 markMCQ
$N(20, 4)$
(a) $(18, 22)$
(b) $(16, 24)$
(c) $(14, 26)$
(d) $(12, 28)$
›Reveal solutionSolution

With $\mu=20$ , $\sigma=\sqrt{4}=2$ , the interval covering $99.73\%$ is $\mu\pm 3\sigma = 20\pm 6 = \mathbf{(14,26)}$ .

Given: $N(20,4)$ means mean $\mu=20$ and variance $\sigma^2=4$ , so $\sigma=2$ .

…

Start your 14-day free trial to see the full solution →
CBSE 2022Set MARCH1 mark
$\mu - 2\sigma$
›Reveal solutionSolution
The area under the normal curve between $\mu-2\sigma$ and $\mu+2\sigma$ is about $\mathbf{95.45\%}$ .

Explanation (empirical rule). For a normal distribution:

$\mu \pm 1\sigma$ covers about $68.27\%$ ,

$\mu \pm 2\sigma$ covers about $95.45\%$ , …
Start your 14-day free trial to see the full solution →

←Previous questionQ1 — In normal distribution usually which limits include 99% of the observa…→Next questionQ3 — For which value of standard normal variable, the standard normal curve…

Q.Mean and the first quartile for a normal distribution are $11$ and $3$ respectively. Which of the following is the value of the third quartile?

Concept understanding — Properties of the Normal Distribution

Q.Mean and the first quartile for a normal distribution are 111111 and 333 respectively. Which of the following is the value of the third quartile?

🎓Unlock everything free for 14 days

Q.Mean and the first quartile for a normal distribution are $11$ and $3$ respectively. Which of the following is the value of the third quartile?

Unlock everything free for 14 days