Q.Find the mean and variance for the first $n$ natural numbers.

Concept & Intuition

The binomial distribution models the number of successes in $n$ independent trials, each with success probability $p$. Its mean is $np$ and variance is $np(1-p)$. When given these two numbers, we can solve for $n$ and $p$ uniquely. Once we know the distribution, computing $P(X \geq 6)$ is just summing the probabilities for $X = 6, 7, 8$ (since $n$ turns out to be small). The trick is to notice that the variance is less than the mean, which forces $p < 1/2$ and gives a neat fraction.

Step-by-step solution

1. Set up equations from given mean and variance Mean: $np = \frac{4}{3}$ Variance: $np(1-p) = \frac{10}{9}$ Divide variance by mean:

So $1-p = \frac{5}{6}$, hence $p = \frac{1}{6}$.

2. Find $n$ From $np = \frac{4}{3}$ and $p = \frac{1}{6}$, we have

So the distribution is $X \sim \text{Binomial}(n=8, p=\frac{1}{6})$.

3. Write the probability mass function

4. Compute $P(X \geq 6)$ This is $P(X=6) + P(X=7) + P(X=8)$.
   • For $k=6$: $\binom{8}{6} = 28$, so

• For $k=7$: $\binom{8}{7} = 8$, so

• For $k=8$: $\binom{8}{8} = 1$, so

5. Sum them

For $X \sim \text{Bin}(n,p)$: mean $= np$, variance $= np(1-p)$.

Find $p$: divide variance by mean,

Find $n$: $np = \dfrac{4}{3} \Rightarrow n\cdot\dfrac{1}{6} = \dfrac{4}{3} \Rightarrow n = 8.$ So $q = 1-p = \dfrac{5}{6}$.

Compute $P(X\ge 6) = P(6)+P(7)+P(8)$:

Adding,

Data: $1, 3, 5, 7, 11, 13, 17, 19, 23$ (nine values).

Mean: $\bar{x}=\dfrac{1+3+5+7+11+13+17+19+23}{9}=\dfrac{99}{9}=11.$

Mean deviation from the mean $M$: the absolute deviations $|x_i-11|$ are

$10, 8, 6, 4, 0, 2, 6, 8, 12$, whose sum is $56$, so

Variance $\sigma^2$: the squared deviations $(x_i-11)^2$ are

$100, 64, 36, 16, 0, 4, 36, 64, 144$, whose sum is $464$, so

Required value:

Step 1 — Find $k$. Total probability is $1$:

So the probabilities are:

Step 2 — Mean.

Step 3 — $E(X^2)$.

Step 4 — Variance.

The mean deviation is a measure of dispersion that tells us, on average, how much the observations in a dataset differ from a central value (in this case, the mean). We use absolute values for the deviations because we are interested in the magnitude of the difference, regardless of whether an observation is above or below the mean. If we didn't take absolute values, the sum of deviations from the mean would always be zero, which wouldn't give us any useful information about spread.

Here's how to calculate it step-by-step:

1. Calculate the mean ($\bar{x}$) of the observations. The given observations are $1, 3, 5, 7, 11, 13, 17, 19, 23$. There are $n = 9$ observations. The sum of observations is: $1 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 = 99$ The mean is:

2. Calculate the absolute deviation ($|x_i - \bar{x}|$) for each observation.

   This means finding the absolute difference between each observation and the mean we just calculated.

   • $|1 - 11| = |-10| = 10$
   • $|3 - 11| = |-8| = 8$
   • $|5 - 11| = |-6| = 6$
   • $|7 - 11| = |-4| = 4$
   • $|11 - 11| = |0| = 0$
   • $|13 - 11| = |2| = 2$
   • $|17 - 11| = |6| = 6$
   • $|19 - 11| = |8| = 8$
   • $|23 - 11| = |12| = 12$

3. Sum the absolute deviations ($\sum |x_i - \bar{x}|$).

   Add all the absolute deviations calculated in the previous step:

4. Calculate the mean deviation from the mean. The mean deviation from the mean is the sum of the absolute deviations divided by the number of observations.

   The mean deviation from the mean is given by:

   $\text{MD}(\bar{x}) = \frac{\sum_{i=1}^{n} |x_i - \bar{x}|}{n}$

   Using our calculated values:

To express this as a mixed fraction:

Concept and Intuition

When we talk about "spread" or "dispersion" in a data set, there are several common measures: variance, standard deviation, and mean deviation. Each captures how far the observations are from the center (usually the mean), but they differ in how they treat the distances.

• Variance averages the squared deviations: $\frac{1}{n}\sum (x_i - \overline{x})^2$.
• Standard deviation is the square root of variance.
• Mean deviation (or mean absolute deviation) averages the absolute deviations: $\frac{1}{n}\sum |x_i - \overline{x}|$.

The question literally asks: "What is the mean of the absolute deviations from $\overline{x}$?" That is exactly the definition of the mean deviation. No calculation is needed — it's a matter of recognizing the terminology.

Step-by-Step Reasoning

1. Identify what is being described The phrase "mean of the absolute deviations of these observations from $\overline{x}$" translates directly to:

This is a standard statistical formula.

2. Match the formula to the options

   • (A) Variance: $\frac{1}{n}\sum (x_i - \overline{x})^2$ — uses squares, not absolute values.
   • (B) Mean proportion: Not a standard measure of dispersion; irrelevant here.
   • (C) Standard deviation: $\sqrt{\frac{1}{n}\sum (x_i - \overline{x})^2}$ — again, squares and a square root.
   • (D) Mean deviation: Exactly $\frac{1}{n}\sum |x_i - \overline{x}|$ — matches perfectly.

3. Confirm by elimination

   The only option that uses absolute values (not squares) is the mean deviation. The other options involve squaring, which changes the nature of the measure (e.g., giving more weight to larger deviations).

The problem asks us to calculate the sum of the mean deviation from the mean ($M_1$) and the mean deviation from the median ($M_2$) for the first 15 even integers.

Concept and Intuition

Mean deviation is a measure of dispersion, indicating the average absolute difference between each data point and a central value (either the mean or the median). It helps us understand how spread out the data points are around that central tendency.

• Mean ($\bar{x}$): The arithmetic average of all observations. It is calculated as the sum of all observations divided by the number of observations.
• Median ($M$): The middle value of a dataset when arranged in ascending or descending order. For an odd number of observations, it is the single middle value.
• Mean Deviation from Mean ($M_1$): This is the average of the absolute differences between each observation and the mean of the dataset.
• Mean Deviation from Median ($M_2$): This is the average of the absolute differences between each observation and the median of the dataset.

For a symmetric distribution, such as an arithmetic progression, the mean and median often coincide. If they do, the mean deviation from the mean and the mean deviation from the median will be identical.

Step-by-Step Calculation

1. Identify the Data Set

   The first 15 even integers are $2, 4, 6, \dots$.

   This is an arithmetic progression with the first term $a = 2$, common difference $d = 2$, and number of terms $n = 15$.

   The last term is $l = a + (n-1)d = 2 + (15-1)2 = 2 + 14 \times 2 = 2 + 28 = 30$.

   So, the data set is $\{2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30\}$.

2. Calculate the Mean ($\bar{x}$)

   The sum of an arithmetic progression is $S_n = \frac{n}{2}(a + l)$.

   $S_{15} = \frac{15}{2}(2 + 30) = \frac{15}{2}(32) = 15 \times 16 = 240$.

   The mean is $\bar{x} = \frac{S_n}{n} = \frac{240}{15} = 16$.

3. Calculate $M_1$ (Mean Deviation from Mean)

   The formula for mean deviation from the mean is:

   $M_1 = \frac{\sum_{i=1}^{n} |x_i - \bar{x}|}{n}$

   Here, $\bar{x} = 16$. We need to find the sum of absolute deviations from 16.

   The deviations are:

   $|2-16|=14$

   $|4-16|=12$

   $|6-16|=10$

   $|8-16|=8$

   $|10-16|=6$

   $|12-16|=4$

   $|14-16|=2$

   $|16-16|=0$

   $|18-16|=2$

   $|20-16|=4$

   $|22-16|=6$

   $|24-16|=8$

   $|26-16|=10$

   $|28-16|=12$

   $|30-16|=14$

   Sum of absolute deviations $\sum |x_i - \bar{x}| = (14+12+10+8+6+4+2) + 0 + (2+4+6+8+10+12+14)$.

   This sum can be written as $2 \times (2+4+6+8+10+12+14)$.

   The sum of an arithmetic progression $2+4+\dots+14$ (7 terms) is $\frac{7}{2}(2+14) = \frac{7}{2}(16) = 7 \times 8 = 56$.

   So, $\sum |x_i - \bar{x}| = 2 \times 56 = 112$.

   Therefore, $M_1 = \frac{112}{15}$.

4. Calculate the Median ($M$)

   Since there are $n=15$ observations (an odd number), the median is the $\left(\frac{n+1}{2}\right)$-th term.

   Median = $\left(\frac{15+1}{2}\right)$-th term = $8$-th term.

   The terms are $2, 4, 6, 8, 10, 12, 14, 16, \dots$.

   The 8th term is $2 \times 8 = 16$.

   So, the median $M = 16$.

5. Calculate $M_2$ (Mean Deviation from Median)

   The formula for mean deviation from the median is:

   $M_2 = \frac{\sum_{i=1}^{n} |x_i - M|}{n}$

   Here, $M = 16$.

   Since the median is 16, which is the same as the mean, the sum of absolute deviations from the median will be identical to the sum of absolute deviations from the mean calculated in step 3.

   $\sum |x_i - M| = \sum |x_i - 16| = 112$.

   Therefore, $M_2 = \frac{112}{15}$.

6. Calculate $M_1 + M_2$

   $M_1 + M_2 = \frac{112}{15} + \frac{112}{15} = \frac{112 + 112}{15} = \frac{224}{15}$.