Easy Practice — Set 1

Quick Recap — Measures of Central Tendency & Dispersion

• Mean $\bar x=\dfrac{\sum x}{n}$; median = middle value (average the middle two if $n$ is even); mode = most frequent.
• Range $=\max-\min$; variance $\sigma^2=\dfrac{\sum(x-\bar x)^2}{n}$; standard deviation $\sigma=\sqrt{\text{variance}}$.
• $\sum(x-\bar x)=0$ always; variance is never negative and is $0$ only for identical values.
• Shortcut: $\sigma^2=\dfrac{\sum x^2}{n}-\bar x^2$.

Beyond-NCERT JEE Formulae

Combined mean (two groups). If group 1 has $n_1$ values with mean $\bar x_1$ and group 2 has $n_2$ values with mean $\bar x_2$, then the mean of the pooled data is

$$\bar x=\dfrac{n_1\bar x_1+n_2\bar x_2}{n_1+n_2}.$$

This is the size-weighted average of the two means, not their plain average.

Combined variance (two groups). With group variances $\sigma_1^2,\sigma_2^2$ and group means $\bar x_1,\bar x_2$,

$$\sigma^2=\dfrac{n_1(\sigma_1^2+d_1^2)+n_2(\sigma_2^2+d_2^2)}{n_1+n_2},\qquad d_1=\bar x_1-\bar x,\ \ d_2=\bar x_2-\bar x,$$

where $\bar x$ is the combined mean. [JEE Tip] Never drop the $d^2$ correction: the scatter of the two group means about the overall mean adds genuine spread, so the pooled variance is usually larger than the weighted average of $\sigma_1^2$ and $\sigma_2^2$.

Total probability theorem. If the events $E_1,E_2,\ldots,E_k$ form a partition of the sample space (mutually exclusive, exhaustive, each of positive probability), then for any event $A$

$$P(A)=\sum_{i=1}^{k}P(E_i)\,P(A\mid E_i).$$

This weighted sum is exactly the denominator that underlies Bayes' theorem.

Mean deviation is least about the median. For any data set the sum $\sum|x-a|$ is minimised when $a$ equals the median; hence the mean deviation about the median is never larger than the mean deviation about the mean. Mean deviation about the mean, by definition, takes $a=\bar x$.

Binomial extras. For $X\sim B(n,p)$ with $q=1-p$: the most likely value (mode) is $\lfloor (n+1)p\rfloor$ when $(n+1)p$ is not an integer, and the "at least one" probability is $P(X\ge 1)=1-P(X=0)=1-q^{\,n}$.

Solved Examples — Beyond-NCERT Formulae

Example 1 — Combined mean. Group A has the $3$ scores $9,12,15$ and Group B has the $2$ scores $13,21$. Find the mean of all $5$ scores.

• Group means: $\bar x_A=\dfrac{9+12+15}{3}=12$ and $\bar x_B=\dfrac{13+21}{2}=17$.
• Combined mean: $\bar x=\dfrac{n_A\bar x_A+n_B\bar x_B}{n_A+n_B}=\dfrac{3(12)+2(17)}{5}=\dfrac{36+34}{5}=14$.
• Pooling check: $\dfrac{9+12+15+13+21}{5}=\dfrac{70}{5}=14$.

Answer: $\bar x=14$.

Example 2 — Combined variance. For the same data, Group A $(9,12,15)$ has variance $\sigma_A^2=6$ and Group B $(13,21)$ has variance $\sigma_B^2=16$. Find the variance of all $5$ scores.

• Deviations of the group means from $\bar x=14$: $d_A=12-14=-2$ and $d_B=17-14=3$.
• Combined variance:

$$\sigma^2=\dfrac{n_A(\sigma_A^2+d_A^2)+n_B(\sigma_B^2+d_B^2)}{n_A+n_B}=\dfrac{3(6+4)+2(16+9)}{5}=\dfrac{30+50}{5}=16.$$

• Pooling check about $14$: the squared deviations $25,4,1,1,49$ sum to $80$, so $\sigma^2=\dfrac{80}{5}=16$.

[JEE Tip] Dropping the $d^2$ terms gives the wrong value $\dfrac{3(6)+2(16)}{5}=10$; the $d^2$ correction restores the missing between-group spread.

Answer: $\sigma^2=16$.

Example 3 — Total probability (two bags). Bag I holds $5$ red and $3$ green balls; Bag II holds $2$ red and $2$ green balls. A bag is chosen at random and one ball is drawn. Find $P(\text{red})$.

• Priors $P(I)=P(II)=\dfrac12$; conditionals $P(R\mid I)=\dfrac{5}{8}$ and $P(R\mid II)=\dfrac{2}{4}=\dfrac12$.
• Total probability: $P(R)=P(I)\,P(R\mid I)+P(II)\,P(R\mid II)=\dfrac12\cdot\dfrac58+\dfrac12\cdot\dfrac12=\dfrac{5}{16}+\dfrac{4}{16}=\dfrac{9}{16}$.

Answer: $P(\text{red})=\dfrac{9}{16}$.

Example 4 — Total probability (three suppliers). Suppliers $S_1,S_2,S_3$ provide $25\%,25\%,50\%$ of a shop's items, with defective rates $4\%,8\%,2\%$ respectively. What fraction of all items are defective?

• The suppliers partition the items: $P(S_1)=0.25$, $P(S_2)=0.25$, $P(S_3)=0.50$, and $P(D\mid S_1)=0.04$, $P(D\mid S_2)=0.08$, $P(D\mid S_3)=0.02$.
• Total probability:

$$P(D)=\sum_i P(S_i)\,P(D\mid S_i)=0.25(0.04)+0.25(0.08)+0.50(0.02)=0.01+0.02+0.01=0.04.$$

Answer: $P(D)=0.04$, i.e. $4\%$ of the items are defective.

Example 5 — Mean deviation: median beats mean. For the data $2,4,6,8,20$, compare the mean deviation about the median with that about the mean.

• Here $n=5$: the median is the middle value $6$, and the mean is $\bar x=\dfrac{2+4+6+8+20}{5}=\dfrac{40}{5}=8$.
• MD about the median: $\dfrac{|2-6|+|4-6|+|6-6|+|8-6|+|20-6|}{5}=\dfrac{4+2+0+2+14}{5}=\dfrac{22}{5}=4.4$.
• MD about the mean: $\dfrac{|2-8|+|4-8|+|6-8|+|8-8|+|20-8|}{5}=\dfrac{6+4+2+0+12}{5}=\dfrac{24}{5}=4.8$.
• The value about the median, $4.4$, is smaller than the value about the mean, $4.8$, exactly as the theory guarantees.

Answer: MD is $4.4$ about the median versus $4.8$ about the mean, so the median gives the smaller mean deviation.

Example 6 — Binomial mode. For $X\sim B\left(10,\dfrac13\right)$, find the most likely number of successes.

• Formula: mode $=\lfloor (n+1)p\rfloor=\left\lfloor 11\cdot\dfrac13\right\rfloor=\left\lfloor\dfrac{11}{3}\right\rfloor=\lfloor 3.67\rfloor=3$.
• Why it works: the consecutive-term ratio is $\dfrac{P(X=k)}{P(X=k-1)}=\dfrac{n-k+1}{k}\cdot\dfrac{p}{q}=\dfrac{11-k}{k}\cdot\dfrac12$, which is $>1$ for $k\le 3$ and $<1$ from $k=4$ onward, so the probabilities rise up to $k=3$ and then fall.

Answer: the most likely value is $X=3$.

Example 7 — At least one. A machine turns out items that are defective independently with probability $p=\dfrac14$. If $3$ items are produced, find the probability that at least one is defective.

• Use the complement with $q=1-\dfrac14=\dfrac34$ and $n=3$: $P(X\ge 1)=1-P(X=0)=1-q^{\,n}$.
• Compute: $P(X\ge 1)=1-\left(\dfrac34\right)^3=1-\dfrac{27}{64}=\dfrac{37}{64}\approx 0.578$.

Answer: $P(X\ge 1)=\dfrac{37}{64}$.