Statistics flash cards

1. 1.Write the formula for variance of grouped/ungrouped data using the direct method.

   $\sigma^2=\dfrac{1}{N}\sum f_i(x_i-\bar{x})^2=\dfrac{1}{N}\sum f_i x_i^2-\left(\dfrac{1}{N}\sum f_i x_i\right)^2$

   Hint: Mean of squares minus square of mean

2. 2.What is the shortcut (step-deviation) method formula for variance when class intervals are equal in size $h$?

   $\sigma^2=\dfrac{h^2}{N^2}\left[N\sum f_i y_i^2-\left(\sum f_i y_i\right)^2\right]$, where $y_i=\dfrac{x_i-A}{h}$

   Hint: Assumed mean $A$ and $y_i$ deviations

3. 3.Define Standard Deviation ($SD$) and Mean Deviation about mean ($M.D.$) in terms of $\sigma$ notation.

   $SD=\sigma=\sqrt{\dfrac{1}{N}\sum f_i(x_i-\bar{x})^2}$; $M.D.(\bar x)=\dfrac{1}{N}\sum f_i|x_i-\bar{x}|$

   Hint: SD squares deviations, MD uses absolute deviations

4. 4.How do variance and standard deviation change if every observation is (a) increased by a constant $k$, (b) multiplied by a constant $k$?

   (a) Variance and $SD$ unchanged; (b) new variance $=k^2\sigma^2$, new $SD=|k|\sigma$

   Hint: Shift vs scale property

5. 5.State the Coefficient of Variation ($C.V.$) formula and its use in comparing two series.

   $C.V.=\dfrac{\sigma}{\bar{x}}\times100\%$; the series with lower $C.V.$ is more consistent (less variable)

   Hint: Relative measure of dispersion, percentage form

6. 6.Group A has $n_1=10$ observations with mean $\bar{x}_1=15$ and variance $\sigma_1^2=8$. Group B has $n_2=15$ observations with mean $\bar{x}_2=20$ and variance $\sigma_2^2=12$. Find the mean and variance of all $25$ observations combined.

   Combined mean $\bar{X}=\dfrac{n_1\bar{x}_1+n_2\bar{x}_2}{n_1+n_2}=\dfrac{10(15)+15(20)}{25}=\dfrac{450}{25}=18$. With $d_1=\bar{x}_1-\bar{X}=-3$ and $d_2=\bar{x}_2-\bar{X}=2$, the combined variance is $\sigma^2=\dfrac{n_1(\sigma_1^2+d_1^2)+n_2(\sigma_2^2+d_2^2)}{n_1+n_2}=\dfrac{10(8+9)+15(12+4)}{25}=\dfrac{170+240}{25}=\dfrac{410}{25}$. So $\bar{X}=18$, $\sigma^2=\dfrac{82}{5}=16.4$.

   Hint: Use the combined mean formula first, then add each group's deviation-squared correction term.

7. 7.Two batches of $50$ students each write a test. Batch 1: mean $25$, variance $4$. Batch 2: mean $35$, variance $9$. Compute the variance of the combined $100$ scores.

   Combined mean $=\dfrac{50(25)+50(35)}{100}=30$. Deviations $d_1=25-30=-5,\ d_2=35-30=5$. Combined variance $=\dfrac{50(4+25)+50(9+25)}{100}=\dfrac{50(29)+50(34)}{100}=\dfrac{1450+1700}{100}=\dfrac{3150}{100}$. So combined variance $=\dfrac{63}{2}=31.5$.

   Hint: Equal group sizes simplify the weighted terms — don't forget the $d_i^2$ correction.

8. 8.A class is split into two sections: Section I has $20$ students with mean marks $10$ and variance $16$; Section II has $30$ students with mean marks $20$ and variance $21$. Find the SD of the entire class of $50$ students.

   Combined mean $\bar X=\dfrac{20(10)+30(20)}{50}=\dfrac{200+600}{50}=16$. Deviations: $d_1=10-16=-6,\ d_2=20-16=4$. Combined variance $=\dfrac{20(16+36)+30(21+16)}{50}=\dfrac{20(52)+30(37)}{50}=\dfrac{1040+1110}{50}=\dfrac{2150}{50}=43$. So SD $=\sqrt{43}\approx6.56$.

   Hint: Combine means first, then use the two-group pooled-variance formula with $d_i^2$.

9. 9.Store A recorded $40$ daily sales with mean $55$ and variance $25$ (in suitable squared units); Store B recorded $60$ days with mean $60$ and variance $36$. Find the combined variance for all $100$ days.

   Combined mean $=\dfrac{40(55)+60(60)}{100}=\dfrac{2200+3600}{100}=58$. Deviations $d_1=55-58=-3,\ d_2=60-58=2$. Combined variance $=\dfrac{40(25+9)+60(36+4)}{100}=\dfrac{40(34)+60(40)}{100}=\dfrac{1360+2400}{100}=\dfrac{3760}{100}$. So combined variance $=\dfrac{188}{5}=37.6$.

   Hint: Weight each group's variance plus squared deviation from the pooled mean.

10. 10.Two equal-sized samples ($n_1=n_2=25$) have identical variance $9$ but means $30$ and $40$ respectively. Find the variance of the pooled sample of $50$ observations.

    Since $n_1=n_2$, combined mean $=\dfrac{30+40}{2}=35$, so $d_1=-5,d_2=5$ (equal in magnitude). Combined variance $=\dfrac{25(9+25)+25(9+25)}{50}=\dfrac{25(34)\times2}{50}=34$. Even though both group variances are only $9$, the gap between the means adds an extra $25$ to the pooled variance, giving combined variance $=34$.

    Hint: Equal group sizes and equal individual variances still leave a mean-gap correction term.

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