110 marksNumericalCoefficient of variationAnswer

What is the difference between relative and absolute statistical measure? J.J. Thomson discovered the electron by isolating negatively charged particles for which he could measure the mass-charge ratio. This ratio appeared to be constant over a wide range of experimental conditions and, consequently, could be characteristics of new particles. His observations from two different cathode-ray tubes that used air as the gas. Which tube is more homogeneous?

$\begin{array}{|c|c|c|c|c|c|c|c|}\hline \text{Tube I} & 0.57 & 0.34 & 0.43 & 0.32 & 0.48 & 0.40 & 0.40 \\ \hline \text{Tube II} & 0.53 & 0.47 & 0.47 & 0.51 & 0.63 & 0.61 & 0.48 \\ \hline \end{array}$

[10]

What is the difference between relative and absolute statistical measure? J.J. Thomson discovered the electron by isolating negatively charged particles for which he could measure the mass-charge ratio. This ratio appeared to be constant over a wide range of experimental conditions and, consequently, could be characteristics of new particles. His observations from two different cathode-ray tubes that used air as the gas. Which tube is more homogeneous?

$$\begin{array}{|c|c|c|c|c|c|c|c|}\hline \text{Tube I} & 0.57 & 0.34 & 0.43 & 0.32 & 0.48 & 0.40 & 0.40 \ \hline \text{Tube II} & 0.53 & 0.47 & 0.47 & 0.51 & 0.63 & 0.61 & 0.48 \ \hline \end{array}$$

[10]

Absolute measures of dispersion: - Expressed in the same units as the original data - Examples: Range, Quartile Deviation, Mean Deviation, Standard Deviation - Measure scatter in absolute terms - Cannot compare variability between series...

210 marksNumericalSimple linear regression modelAnswer

Pumpkin Weight Prediction Analysis

In Nepal, during the fall harvest season pumpkins are sold in large quantities at farm stands. Often instead of weighing the pumpkin prior to sale, the farm stand operators will just place the pumpkin in the appropriate circular cutout on the counter. When asked why this was done, one farmer replied, "I can tell the weight of the pumpkin from its circumference". To determine whether this was really true, a sample of 8 pumpkins' circumference and weights were measured:

Circumference (cm)	50	55	54	52	37	52	53	47
Weight (gram × 100)	12	20	15	17	50	10	15	14

i) Develop the estimating linear regression equation of weight of the pumpkin to the circumference.

ii) Estimate the weight of pumpkin if the circumference value is 40 cm. Can it be possible to predict the weight of a pumpkin if its circumference is 80 cm?

[10]

Circumference X (cm) Weight Y (×100 g) ------ 50 12 55 20 54 15 52 17 37 50 52 10 53 15 47 14 - n = 8 - Required: (i) regression of weight on circumference, (ii) estimate weight at X = 40, and comment on X = 80. Computation table: X Y XY...

310 marksNumericalNormal distribution properties and applicaAnswer

Define normal distribution and standard normal distribution. In an examination, 10% students got less than 20 marks and 95% got less than 75 marks. Assuming the distribution to be normal, find the mean and standard deviation.[10]

Model Answer: Normal Distribution and Standard Normal Distribution

Given Data

$P(X < 20) = 0.10$ (10% got less than 20 marks)
$P(X < 75) = 0.95$ (95% got less than 75 marks)
Distribution is normal with unknown mean $\mu$ and standard deviation $\sigma$

Part 1: Definitions

Normal Distribution

A normal distribution is a continuous probability distribution that is symmetric and bell-shaped about its mean. It is completely specified by two parameters: the mean $\mu$ and the standard deviation $\sigma$ . Its probability density function is:

$f(x) = \frac{1}{\sigma\sqrt{2\pi}}\, e^{-\frac{(x-\mu)^2}{2\sigma^2}}, \quad -\infty < x < \infty$

Key properties:

Mean = Median = Mode
About 68% of data lies within $\mu \pm \sigma$
About 95% within $\mu \pm 2\sigma$
About 99.7% within $\mu \pm 3\sigma$

Standard Normal Distribution

A standard normal distribution is a normal distribution with mean $\mu = 0$ and standard deviation $\sigma = 1$ , denoted $N(0,1)$ . Its density is:

$\phi(z) = \frac{1}{\sqrt{2\pi}}\, e^{-\frac{z^2}{2}}$

Any normal variable $X$ is transformed to standard normal via:

$Z = \frac{X - \mu}{\sigma}$

Part 2: Finding Mean and Standard Deviation

Step 1: Standardize both conditions

$Z_1 = \frac{20 - \mu}{\sigma}, \quad P(Z < Z_1) = 0.10$ $Z_2 = \frac{75 - \mu}{\sigma}, \quad P(Z < Z_2) = 0.95$

Step 2: Read Z-values from tables

$P(Z < Z_1) = 0.10$ (below the mean) gives $Z_1 = -1.28$
$P(Z < Z_2) = 0.95$ gives $Z_2 = 1.645$

Step 3: Form equations

$20 - \mu = -1.28\,\sigma \quad \text{...(1)}$ $75 - \mu = 1.645\,\sigma \quad \text{...(2)}$

Step 4: Solve for $\sigma$

Subtract (1) from (2):

$(75 - \mu) - (20 - \mu) = 1.645\sigma - (-1.28\sigma)$ $55 = 2.925\,\sigma$ $\sigma = \frac{55}{2.925} = 18.80 \text{ marks}$

Step 5: Solve for $\mu$

From equation (1): $\mu = 20 + 1.28\sigma = 20 + 1.28 \times 18.80 = 20 + 24.06 = 44.06 \text{ marks}$

Verification using (2): $75 - 44.06 = 30.94 \quad \text{and} \quad 1.645 \times 18.80 = 30.93 \;\checkmark$

Final Answer

Mean $\mu \approx 44.06$ marks (≈ 44 marks)
Standard Deviation $\sigma \approx 18.80$ marks (≈ 19 marks)

(Note: If $Z_1 = -1.28$ is taken more precisely as $-1.2816$ and $Z_2 = 1.645$ , results are essentially the same: $\sigma \approx 18.79$ , $\mu \approx 44.08$ .)

Model Answer: Normal Distribution and Standard Normal Distribution

Given Data

$P(X < 20) = 0.10$ (10% got less than 20 marks)
$P(X < 75) = 0.95$ (95% got less than 75 marks)
Distribution is normal with unknown mean $\mu$ and standard deviation $\sigma$

Part 1: Definitions

Normal Distribution

A normal distribution is a continuous probability distribution that is symmetric and bell-shaped about its mean. It is completely specified by two parameters: the mean $\mu$ and the standard deviation $\sigma$. Its probability density function is:

$$f(x) = \frac{1}{\sigma\sqrt{2\pi}}, e^{-\frac{(x-\mu)^2}{2\sigma^2}}, \quad -\infty < x < \infty$$

Key properties:

Mean = Median = Mode
About 68% of data lies within $\mu \pm \sigma$
About 95% within $\mu \pm 2\sigma$
About 99.7% within $\mu \pm 3\sigma$

Standard Normal Distribution

A standard normal distribution is a normal distribution with mean $\mu = 0$ and standard deviation $\sigma = 1$, denoted $N(0,1)$. Its density is:

$$\phi(z) = \frac{1}{\sqrt{2\pi}}, e^{-\frac{z^2}{2}}$$

Any normal variable $X$ is transformed to standard normal via:

$$Z = \frac{X - \mu}{\sigma}$$

Part 2: Finding Mean and Standard Deviation

Step 1: Standardize both conditions

$$Z_1 = \frac{20 - \mu}{\sigma}, \quad P(Z < Z_1) = 0.10$$ $$Z_2 = \frac{75 - \mu}{\sigma}, \quad P(Z < Z_2) = 0.95$$

Step 2: Read Z-values from tables

$P(Z < Z_1) = 0.10$ (below the mean) gives $Z_1 = -1.28$
$P(Z < Z_2) = 0.95$ gives $Z_2 = 1.645$

Step 3: Form equations

$$20 - \mu = -1.28,\sigma \quad \text{...(1)}$$ $$75 - \mu = 1.645,\sigma \quad \text{...(2)}$$

Step 4: Solve for $\sigma$

Subtract (1) from (2):

$$(75 - \mu) - (20 - \mu) = 1.645\sigma - (-1.28\sigma)$$ $$55 = 2.925,\sigma$$ $$\sigma = \frac{55}{2.925} = 18.80 \text{ marks}$$

Step 5: Solve for $\mu$

From equation (1): $$\mu = 20 + 1.28\sigma = 20 + 1.28 \times 18.80 = 20 + 24.06 = 44.06 \text{ marks}$$

Verification using (2): $$75 - 44.06 = 30.94 \quad \text{and} \quad 1.645 \times 18.80 = 30.93 ;\checkmark$$

Final Answer

Mean $\mu \approx 44.06$ marks (≈ 44 marks)
Standard Deviation $\sigma \approx 18.80$ marks (≈ 19 marks)

(Note: If $Z_1 = -1.28$ is taken more precisely as $-1.2816$ and $Z_2 = 1.645$, results are essentially the same: $\sigma \approx 18.79$, $\mu \approx 44.08$.)

45 marksPrimary and secondary data sourcesAnswer

What do you mean by secondary data? Write the source of secondary data with special focus to Nepalese context. [5]

Secondary data refers to information that has already been collected, processed, and published by other researchers, organizations, or institutions for purposes other than the current research study. It is data that exists in compiled or...

55 marksNumericalHistogram constructionAnswer

Construct histogram, frequency polygon and frequency curve of the following frequency distribution.

Circumference (cm)	50	55	54	52	37	52	53	47
Weight (gram) '00'	12	20	15	17	50	10	15	14

[5]

The data is presented as paired values of Circumference (cm) and Weight (gram, in hundreds i.e. '00'): Circumference (cm) 50 55 54 52 37 52 53 47 --------------------------- Weight (gram '00') 12 20 15 17 50 10 15 14 Important interpreta...

65 marksNumericalSpearman rank correlation coefficientAnswer

Following data represent the preference of 10 students studying BIT towards two brands of computers namely DELL and HP. Apply Pearson's rank correlation coefficient and compute correlation coefficient. Also interpret your result.

Computer	Student preference
DELL	5	2	9	8	1	10	3	4	6	7
HP	10	5	1	3	8	6	2	7	9	4

[5]

Student 1 2 3 4 5 6 7 8 9 10 ----------------------------------------- DELL (X) 5 2 9 8 1 10 3 4 6 7 HP (Y) 10 5 1 3 8 6 2 7 9 4 Number of students $n = 10$ . Both rows are already ranks (1 to 10, no ties). Step 1: Compute rank difference...

Student 1 2 3 4 5 6 7 8 9 10 ----------------------------------------- DELL (X) 5 2 9 8 1 10 3 4 6 7 HP (Y) 10 5 1 3 8 6 2 7 9 4 Number of students $n = 10$. Both rows are already ranks (1 to 10, no ties). Step 1: Compute rank difference...

75 marksNumericalRandom variables and probability functionsAnswer

Define Random Variable and Solve

Define random variable: A random variable is a function that assigns numerical values to the outcomes of a random experiment. It can be discrete (taking specific values) or continuous (taking any value in an interval).

Given the probability distribution:

$\begin{array}{|c|c|c|c|c|c|}\hline X & -2 & -1 & 0 & 1 & 2 \\ \hline P(X) & 0.2 & K & 0.4 & 2K & K \\ \hline \end{array}$

Define Random Variable and Solve

Define random variable: A random variable is a function that assigns numerical values to the outcomes of a random experiment. It can be discrete (taking specific values) or continuous (taking any value in an interval).

Given the probability distribution:

$$\begin{array}{|c|c|c|c|c|c|}\hline X & -2 & -1 & 0 & 1 & 2 \ \hline P(X) & 0.2 & K & 0.4 & 2K & K \ \hline \end{array}$$

A random variable is a real-valued function that assigns a numerical value to each outcome in the sample space of a random experiment. In other words, it maps the outcomes of a random phenomenon to numbers. Random variables are classifie...

85 marksNumericalBinomial distributionAnswer

During one stage in the manufacture of integrated circuit chips, a coating must be applied. If 70% of chips received a thick enough coating, find the probability that among 15 chips (i) at least 12 will have thick enough coatings, (ii) at most 2 will have thick enough coatings, and (ii) exactly 10 will have thick enough coatings. [5]

Model Answer: Binomial Probability

STEP 1 - Given Data

Probability a chip receives thick enough coating: $p = 0.70$
Failure probability: $q = 1 - p = 0.30$
Number of chips: $n = 15$
$X$ = number of chips with thick enough coating $\sim \text{Binomial}(15, 0.70)$

Formula: $P(X = k) = \binom{15}{k}(0.70)^k (0.30)^{15-k}$

STEP 2 - Solve

(i) $P(X \ge 12)$

$P(X \ge 12) = P(12) + P(13) + P(14) + P(15)$

P(X=12): $\binom{15}{12}=455$ $455 \times (0.70)^{12}\times(0.30)^3 = 455 \times 0.0138413 \times 0.027 = 0.17004$

Let me compute carefully: $(0.70)^{12} = 0.01384129$ , $(0.30)^3 = 0.027$ $455 \times 0.01384129 \times 0.027 = 0.170043$

P(X=13): $\binom{15}{13}=105$ , $(0.70)^{13}=0.00968890$ , $(0.30)^2=0.09$ $105 \times 0.00968890 \times 0.09 = 0.091560$

P(X=14): $\binom{15}{14}=15$ , $(0.70)^{14}=0.00678223$ , $(0.30)^1=0.30$ $15 \times 0.00678223 \times 0.30 = 0.030520$

P(X=15): $\binom{15}{15}=1$ , $(0.70)^{15}=0.00474756$ $1 \times 0.00474756 = 0.004748$

$P(X \ge 12) = 0.170043 + 0.091560 + 0.030520 + 0.004748 = 0.29687$

$\boxed{P(X \ge 12) \approx 0.2969}$

(ii) $P(X \le 2)$

P(X=0): $(0.30)^{15} = 1.435 \times 10^{-8} \approx 0.0000$

P(X=1): $15 \times 0.70 \times (0.30)^{14} = 15 \times 0.70 \times 4.7830\times10^{-8} \approx 5.02\times10^{-7} \approx 0.0000$

P(X=2): $105 \times (0.70)^2 \times (0.30)^{13} = 105 \times 0.49 \times 1.5943\times10^{-7} \approx 8.20\times10^{-6} \approx 0.0000$

$P(X \le 2) \approx 0.0000088 \approx 0.00001$

$\boxed{P(X \le 2) \approx 0.00001 \text{ (essentially negligible)}}$

(iii) $P(X = 10)$

$\binom{15}{10} = 3003$ , $(0.70)^{10} = 0.0282475$ , $(0.30)^5 = 0.00243$ $P(X=10) = 3003 \times 0.0282475 \times 0.00243 = 0.20613$

$\boxed{P(X = 10) \approx 0.2061}$

Summary

(i) $P(X \ge 12) \approx 0.2969$ (29.69%)
(ii) $P(X \le 2) \approx 0.00001$ (negligible)
(iii) $P(X = 10) \approx 0.2061$ (20.61%)

Check the powers of 0.70: writing $(0.70)^{12}=0.0317$ , which is actually $(0.70)^{10}$ , inflates the sum in part (i) to 0.6901. The correct value is 0.2969; parts (ii) and (iii) are unaffected.

Model Answer: Binomial Probability

STEP 1 - Given Data

Probability a chip receives thick enough coating: $p = 0.70$
Failure probability: $q = 1 - p = 0.30$
Number of chips: $n = 15$
$X$ = number of chips with thick enough coating $\sim \text{Binomial}(15, 0.70)$

Formula: $$P(X = k) = \binom{15}{k}(0.70)^k (0.30)^{15-k}$$

STEP 2 - Solve

(i) $P(X \ge 12)$

$P(X \ge 12) = P(12) + P(13) + P(14) + P(15)$

P(X=12): $\binom{15}{12}=455$ $$455 \times (0.70)^{12}\times(0.30)^3 = 455 \times 0.0138413 \times 0.027 = 0.17004$$

Let me compute carefully: $(0.70)^{12} = 0.01384129$, $(0.30)^3 = 0.027$ $$455 \times 0.01384129 \times 0.027 = 0.170043$$

P(X=13): $\binom{15}{13}=105$, $(0.70)^{13}=0.00968890$, $(0.30)^2=0.09$ $$105 \times 0.00968890 \times 0.09 = 0.091560$$

P(X=14): $\binom{15}{14}=15$, $(0.70)^{14}=0.00678223$, $(0.30)^1=0.30$ $$15 \times 0.00678223 \times 0.30 = 0.030520$$

P(X=15): $\binom{15}{15}=1$, $(0.70)^{15}=0.00474756$ $$1 \times 0.00474756 = 0.004748$$

$$P(X \ge 12) = 0.170043 + 0.091560 + 0.030520 + 0.004748 = 0.29687$$

$$\boxed{P(X \ge 12) \approx 0.2969}$$

(ii) $P(X \le 2)$

P(X=0): $(0.30)^{15} = 1.435 \times 10^{-8} \approx 0.0000$

P(X=1): $15 \times 0.70 \times (0.30)^{14} = 15 \times 0.70 \times 4.7830\times10^{-8} \approx 5.02\times10^{-7} \approx 0.0000$

P(X=2): $105 \times (0.70)^2 \times (0.30)^{13} = 105 \times 0.49 \times 1.5943\times10^{-7} \approx 8.20\times10^{-6} \approx 0.0000$

$$P(X \le 2) \approx 0.0000088 \approx 0.00001$$

$$\boxed{P(X \le 2) \approx 0.00001 \text{ (essentially negligible)}}$$

(iii) $P(X = 10)$

$\binom{15}{10} = 3003$, $(0.70)^{10} = 0.0282475$, $(0.30)^5 = 0.00243$ $$P(X=10) = 3003 \times 0.0282475 \times 0.00243 = 0.20613$$

$$\boxed{P(X = 10) \approx 0.2061}$$

Summary

(i) $P(X \ge 12) \approx 0.2969$ (29.69%)
(ii) $P(X \le 2) \approx 0.00001$ (negligible)
(iii) $P(X = 10) \approx 0.2061$ (20.61%)

Check the powers of 0.70: writing $(0.70)^{12}=0.0317$, which is actually $(0.70)^{10}$, inflates the sum in part (i) to 0.6901. The correct value is 0.2969; parts (ii) and (iii) are unaffected.

95 marksNumericalBasic probability concepts and rulesAnswer

A new computer virus can enter the system through e-mail or through the internet. There is a 30% chance of receiving this virus hrough e-mail. There is a 40% chance of receiving it through the internet. Also, the virus enters the system simultaneously through e-mail and the internet with probability 0.15. What is, the probability that the virus does not enter the system at all? [5]

$P(E)$ = probability virus enters through e-mail = $0.30$ - $P(I)$ = probability virus enters through internet = $0.40$ - $P(E \cap I)$ = probability virus enters through both simultaneously = $0.15$ Step 1: Probability that the virus ...

$P(E)$ = probability virus enters through e-mail = $0.30$ - $P(I)$ = probability virus enters through internet = $0.40$ - $P(E \cap I)$ = probability virus enters through both simultaneously = $0.15$ Step 1: Probability that the virus ...

105 marksNumericalCoefficient of kurtosisAnswer

The standard deviation of a symmetric distribution is 9. Compute the possible value of fourth central moment for the distribution to be (i) mesokurtic (ii) platykurtic, and (iii) leptokurtic. [5]

Model Answer: Fourth Central Moment and Kurtosis

STEP 1 - Given Data

Standard deviation: $\sigma = 9$
Distribution is symmetric
Required: fourth central moment $\mu_4$ for the distribution to be (i) mesokurtic, (ii) platykurtic, (iii) leptokurtic

STEP 2 - Solution

Key relation (coefficient of kurtosis):

$\beta_2 = \frac{\mu_4}{\mu_2^2} = \frac{\mu_4}{\sigma^4}$

so $\mu_4 = \beta_2 \cdot \sigma^4$ .

Compute $\sigma^4$ :

$\sigma^2 = 9^2 = 81$ $\sigma^4 = 81^2 = 6561$

Kurtosis criteria:

Mesokurtic: $\beta_2 = 3$
Platykurtic: $\beta_2 < 3$
Leptokurtic: $\beta_2 > 3$

(i) Mesokurtic ( $\beta_2 = 3$ )

$\mu_4 = 3 \times 6561 = 19683$

(ii) Platykurtic ( $\beta_2 < 3$ )

$\mu_4 < 3 \times 6561$ $\mu_4 < 19683$ (e.g. $\mu_4 = 15000$ )

(iii) Leptokurtic ( $\beta_2 > 3$ )

$\mu_4 > 3 \times 6561$ $\mu_4 > 19683$ (e.g. $\mu_4 = 25000$ )

Summary

Type	Condition	Fourth Central Moment $\mu_4$
Mesokurtic	$\beta_2 = 3$	$= 19683$
Platykurtic	$\beta_2 < 3$	$< 19683$
Leptokurtic	$\beta_2 > 3$	$> 19683$

Model Answer: Fourth Central Moment and Kurtosis

STEP 1 - Given Data

Standard deviation: $\sigma = 9$
Distribution is symmetric
Required: fourth central moment $\mu_4$ for the distribution to be (i) mesokurtic, (ii) platykurtic, (iii) leptokurtic

STEP 2 - Solution

Key relation (coefficient of kurtosis):

$$\beta_2 = \frac{\mu_4}{\mu_2^2} = \frac{\mu_4}{\sigma^4}$$

so $\mu_4 = \beta_2 \cdot \sigma^4$.

Compute $\sigma^4$:

$$\sigma^2 = 9^2 = 81$$ $$\sigma^4 = 81^2 = 6561$$

Kurtosis criteria:

Mesokurtic: $\beta_2 = 3$
Platykurtic: $\beta_2 < 3$
Leptokurtic: $\beta_2 > 3$

(i) Mesokurtic ($\beta_2 = 3$)

$$\mu_4 = 3 \times 6561 = 19683$$

(ii) Platykurtic ($\beta_2 < 3$)

$$\mu_4 < 3 \times 6561$$ $$\mu_4 < 19683$$ (e.g. $\mu_4 = 15000$)

(iii) Leptokurtic ($\beta_2 > 3$)

$$\mu_4 > 3 \times 6561$$ $$\mu_4 > 19683$$ (e.g. $\mu_4 = 25000$)

Summary

Type	Condition	Fourth Central Moment $\mu_4$
Mesokurtic	$\beta_2 = 3$	$= 19683$
Platykurtic	$\beta_2 < 3$	$< 19683$
Leptokurtic	$\beta_2 > 3$	$> 19683$

115 marksNumericalConfidence interval for population meanAnswer

The mass of vitamin E in a capsule manufactured by a certain drug company is normally distributed with standard deviation 0.042 mg. A random sample of 36 capsules was analyzed and the mean mass of vitamin E was found to be 5.12 mg. Calculate a symmetric 95% confidence interval for the population mean mass of vitamin E per capsule. [5]

Population standard deviation: $\sigma = 0.042$ mg (known) - Sample size: $n = 36$ - Sample mean: $\bar{x} = 5.12$ mg - Confidence level: $95\%$ - Distribution: Normal Since $\sigma$ is known and the population is normal, use the $z$ -d...

Population standard deviation: $\sigma = 0.042$ mg (known) - Sample size: $n = 36$ - Sample mean: $\bar{x} = 5.12$ mg - Confidence level: $95%$ - Distribution: Normal Since $\sigma$ is known and the population is normal, use the $z$-d...

125 marksBox and whisker plotsAnswer

Write short notes on : a.) Box and whisker plot. Write short notes on : b.) Census survey and sample survey [2.5+2.5]

A box and whisker plot (or boxplot) is a graphical method for displaying the distribution and spread of numerical data based on five-number summary statistics. Components: - Minimum value: The smallest data point (or lower whisker endpoi...

Basic Statistics 2079 question paper

Model Answer: Normal Distribution and Standard Normal Distribution

Given Data

Part 1: Definitions

Normal Distribution

Standard Normal Distribution

Part 2: Finding Mean and Standard Deviation

Final Answer

Model Answer: Normal Distribution and Standard Normal Distribution

Given Data

Part 1: Definitions

Normal Distribution

Standard Normal Distribution

Part 2: Finding Mean and Standard Deviation

Final Answer

Model Answer: Binomial Probability

STEP 1 - Given Data

STEP 2 - Solve

(i) P(X≥12)P(X \ge 12)P(X≥12)

(ii) P(X≤2)P(X \le 2)P(X≤2)

(iii) P(X=10)P(X = 10)P(X=10)

Summary

Model Answer: Binomial Probability

STEP 1 - Given Data

STEP 2 - Solve

(i) $P(X \ge 12)$

(ii) $P(X \le 2)$

(iii) $P(X = 10)$

Summary

Model Answer: Fourth Central Moment and Kurtosis

STEP 1 - Given Data

STEP 2 - Solution

(i) Mesokurtic (β2=3\beta_2 = 3β2​=3)

(ii) Platykurtic (β2<3\beta_2 < 3β2​<3)

(iii) Leptokurtic (β2>3\beta_2 > 3β2​>3)

Summary

Model Answer: Fourth Central Moment and Kurtosis

STEP 1 - Given Data

STEP 2 - Solution

(i) Mesokurtic ($\beta_2 = 3$)

(ii) Platykurtic ($\beta_2 < 3$)

(iii) Leptokurtic ($\beta_2 > 3$)

Summary

Model Answer: Normal Distribution and Standard Normal Distribution

Given Data

Part 1: Definitions

Normal Distribution

Standard Normal Distribution

Part 2: Finding Mean and Standard Deviation

Final Answer

Model Answer: Binomial Probability

STEP 1 - Given Data

STEP 2 - Solve

(i) P(X≥12)P(X \ge 12)P(X≥12)

(ii) P(X≤2)P(X \le 2)P(X≤2)

(iii) P(X=10)P(X = 10)P(X=10)

Summary

Model Answer: Fourth Central Moment and Kurtosis

STEP 1 - Given Data

STEP 2 - Solution

(i) Mesokurtic (β2=3\beta_2 = 3β2​=3)

(ii) Platykurtic (β2<3\beta_2 < 3β2​<3)

(iii) Leptokurtic (β2>3\beta_2 > 3β2​>3)

Summary

(i) $P(X \ge 12)$

(ii) $P(X \le 2)$

(iii) $P(X = 10)$

(i) Mesokurtic ( $\beta_2 = 3$ )

(ii) Platykurtic ( $\beta_2 < 3$ )

(iii) Leptokurtic ( $\beta_2 > 3$ )

(i) $P(X \ge 12)$

(ii) $P(X \le 2)$

(iii) $P(X = 10)$

(i) Mesokurtic ( $\beta_2 = 3$ )

(ii) Platykurtic ( $\beta_2 < 3$ )

(iii) Leptokurtic ( $\beta_2 > 3$ )