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Maths 3 Marks Question

Statistics · CBSE STD 11 Science (CBSE - English Medium). 8 questions.

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3 Marks Question

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Question 13 Marks
For the frequency distribution:
$x$
$2$
$3$
$4$
$5$
$6$
$7$
$f$
$4$
$9$
$16$
$14$
$11$
$6$
Find the standard distribution.
Answer
$x_i$
$f_i$
$f_ix_i$
$f_ix_i^2$
$2$
$4$
$8$
$16$
$3$
$9$
$27$
$81$
$4$
$16$
$64$
$256$
$5$
$14$
$70$
$350$
$6$
$11$
$66$
$396$
$7$
$6$
$42$
$294$
 
$N = 60$
$\sum\text{f}_\text{i}\text{x}_\text{i}=277$
 $\sum\text{f}_\text{i}\text{x}_\text{i}^2=1393$
$\therefore\ \text{SD}\sqrt{\frac{\sum\text{f}_\text{i}\text{x}^2_\text{i}}{\text{N}}-\Big(\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\text{N}}\Big)^2}$
$=\sqrt{\frac{1393}{60}-\Big(\frac{277}{60}\Big)^2}$
$=\sqrt{23.23-(4.62)^2}$
$=\sqrt{23.21-21.34}$
$=\sqrt{1.87}=1.37$ Hence, the required $ \text{SD}= 1.37$
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Question 23 Marks
Find the mean and variance of the frequency distribution given below:
$\text{x}$ $1\leq\text{x}<3$ $3\leq\text{x}<5$ $5\leq\text{x}<7$ $7\leq\text{x}<10$
$\text{f}$ $6$ $4$ $5$ $1$
Answer
$X$
$f_i$
$X_i$
$f_ix_i$
$f_ix_i^2$
$1-3$
$6$
$2$
$12$
$24$
$3-5$
$4$
$4$
$16$
$64$
$5-7$
$5$
$6$
$30$
$180$
$7-10$
$1$
$8.5$
$8.5$
$72.25$
$Total$
$n = 16$
 
$\sum\text{f}_\text{i}\text{x}_\text{i}=66.5$
$\sum\text{f}_\text{i}\text{x}_\text{i}^2=340.25$
$\therefore\ \text{Mean }=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}=\frac{66.5}{16}=4.13$
And $\text{variance}=\sigma^2=\frac{\sum\text{f}_\text{i}\text{x}^2_\text{i}}{\sum\text{f}_\text{i}}-\Big(\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}\Big)^2$
$=\frac{340.25}{16}-(4.13)^2$
$=21.2656-17.0569=4.21$
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Question 33 Marks
If for a distribution $\sum(\text{x}-5)=3,\sum(\text{x}-5)^2=43$ and the total number of item is 18, find the mean and standard deviation.
Answer
Given, $\text{n}=18,\ \sum(\text{x}-5)=3,\ \sum(\text{x}-5)^2=43$
$\therefore\ \text{Mean}=\text{A}+\frac{\sum(\text{x}-5)}{18}=5+\frac{3}{18}$
$=5+0.1666=5.1666=5.17$
and $\text{SD}=\sqrt{\frac{\sum(\text{x}-5)^2}{\text{n}}-\Big(\frac{\sum(\text{x}-5)}{\text{n}}\Big)^2}$
 $=\sqrt{\frac{43}{18}-\Big(\frac{3}{18}\Big)^2}$
$=\sqrt{2.3889-(0.166)^2}$
$=\sqrt{2.3889-0.0277}$
$=1.53$
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Question 43 Marks
Two sets each of 20 observations, have the same standard derivation 5. The first set has a mean 17 and the second a mean 22. Determine the standard deviation of the set obtained by combining the given two sets.
Answer
Given that $\text{n}_1=20,\sigma_1=5,\bar{\text{x}}_1=17$
and $\text{n}_2=20,\sigma_2=5,\bar{\text{x}}_2=22$
Now we know for combined two series that
$\sigma=\sqrt{\frac{\text{n}_1\text{s}^2_1+\text{n}_2\text{s}^2_2}{\text{n}_1+\text{n}_2}+\frac{\text{n}_1\text{n}_2(\bar{\text{x}}_1-\bar{\text{x}}_2)^2}{(\text{n}_1+\text{n}_2)^2}}$
$=\sqrt{\frac{20\times(5)^2+20\times(5)^2}{20+20}+\frac{20\times20(17-22)^2}{(20+20)^2}}$
$=\sqrt{\frac{1000}{40}+\frac{400\times25}{1600}}$
$=\sqrt{25+\frac{25}{4}}=\sqrt{\frac{125}{4}}$
$=\sqrt{31.25}=5.59$
Hence, the required SD = 5.59
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Question 53 Marks
Mean and standard deviation of 100 items are 50 and 4, respectively. Find the sum of all the item and the sum of the squares of the items.
Answer
Given that $\bar{\text{x}}=50,\text{ n}=100$ and $\text{SD}(\sigma)=4$
$\bar{\text{x}}=\frac{\sum\text{x}_\text{i}}{\text{N}}$
$\Rightarrow50=\frac{\sum\text{x}_\text{i}}{100}$
$\Rightarrow{\sum\text{x}_\text{i}}=5000$
and variance $\sigma^2=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}^2}{\text{N}}-\Big(\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\text{N}}\Big)^2$
$(4)^2=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}^2}{100}=-(50)^2$
$\Rightarrow16=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}^2}{100}-2500$
$\therefore\ \sum\text{f}_\text{i}\text{x}_\text{i}^2=(2500+16)\times100$
$\Rightarrow\sum\text{f}_\text{i}\text{x}_\text{i}^2=2516\times100=251600$
Hence, the required sum are 5000 and 251600
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Question 63 Marks
Find the mean deviation about the median of the following distribution:
Marks obtained
$10$
$11$
$12$
$14$
$15$
No. of students
$2$
$3$
$8$
$3$
$4$ 
Answer
$\text{Marks obtained}$
$f_i$
$cf$
$di = |x_i - M_e|$
$f_id_i$
$10$
$2$
$2$
$2$
$4$
$11$
$3$
$5$
$1$
$3$
$12$
$8$
$13$
$0$
$0$
$14$
$3$
$16$
$2$
$6$
$15$
$4$
$20$
$3$
$12$
$Total$
 $\sum\text{f}_\text{i}=20$     $\sum\text{f}_\text{i}\text{d}_\text{i}=25$
Now, $\text{M}_\text{e}=\Big(\frac{20+1}{2}\Big)\text{item}=\Big(\frac{21}{2}\Big)=10.5^{\text{th}}\text{ item}$
$\therefore\ \text{M}_\text{e}=12$
$\therefore\ \text{MD}=\frac{\sum\text{f}_\text{i}\text{d}_\text{i}}{\sum\text{f}_\text{i}}=\frac{25}{20}=1.25$
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Question 73 Marks
Find the mean deviation about the mean of the distribution:
Size
$20$
$21$
$22$
$23$
$24$
Frequency
$6$
$4$
$5$
$1$
$4$
Answer
$Size(x_i)$
$Frequency(f_i)$
$f_ix_i$
$\text{d}_{\text{i}}=|\text{x}_\text{i}-\bar{\text{x}}|$
$f_id_i$
$20$
$6$
$120$
$1.65$
$9.90$
$21$
$4$
$84$
$0.65$
$2.60$
$22$
$5$
$110$
$0.35$
$1.75$
$23$
$1$
$23$
$1.35$
$1.75$
$24$
$4$
$96$
$2.35$
$9.40$
$Total$
$20$
$433$
$6.35$
$25.00$
Mean $\bar{\text{x}}=\frac{\sum\text{f}_\text{i}\text{x}_\text{i}}{\sum\text{f}_\text{i}}=\frac{433}{20}=21.65$
Mean deviation $\text{MD}=\frac{\sum\text{f}_\text{i}\text{d}_\text{i}}{\sum\text{f}_\text{i}}=\frac{25}{20}=1.25$
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Question 83 Marks
Calculate the mean deviation about the mean of the set of first n natural numbers when n is an even number.
Answer
Consider first n natural number, when n is even i.e., 1, 2, 3, 4 ..... n.
$\therefore\text{ Mean }\bar{\text{x}}=\frac{1+2+3+\ ......\ +\text{n}}{\text{n}}=\frac{\text{n}(\text{n}+1)}{2\text{n}}=\frac{\text{n}+1}{2}$
$\text{MD}=\frac{1}{\text{n}}\Big[\Big|1-\frac{\text{n}+1}{2}\Big|+\Big|2-\frac{\text{n}+1}{2}\Big|+\Big|3-\frac{\text{n}+1}{2}\Big|\Big]+\Big|\frac{\text{n}-2}{2}-\frac{\text{n}+1}{2}\Big|\\+\Big|\frac{\text{n}}{2}-\frac{\text{n}+1}{2}\Big|+\Big|\frac{\text{n}+2}{2}-\frac{\text{n}+1}{2}\Big|+\ .....\ +\Big|\text{n}-\frac{\text{n}+1}{2}\Big|$
$=\frac{1}{\text{n}}\Big[\Big|\frac{\text{n}-1}{2}\Big|+\Big|\frac{3\text{n}-\text{n}}{2}\Big|+\Big|\frac{5-\text{n}}{2}\Big|+\ .....\ +\Big|\frac{-3}{2}\Big|+\Big|\frac{1}{2}\Big|+\ ....\ +\Big|\frac{\text{n}-1}{2}\Big|\Big]$
$=\frac{2}{\text{n}}\Big[\frac{1}{2}+\frac{3}{2}+\ ....\ +\frac{\text{n}-1}{2}\Big]\Big(\frac{\text{n}}{2}\Big)\text{terms}$
$=\frac{1}{\text{n}}\cdot\Big(\frac{\text{n}}{2}\Big)^2$ $\big[\because$ Sum of first n natural numbers $=\text{n}^2\big]$
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3 Marks Question - Maths STD 11 Science Questions - Vidyadip