# Activity- statistics problems

(Redirected from Class10 statistics problems)

# Problem 1

If n =10, ${\displaystyle {\bar {x}}}$ = 12 and
${\displaystyle \sum {x^{2}}=1530findthestandarddeviation}$

# Solution:

${\displaystyle {\frac {\sum x}{n}}={\bar {x}}}$

${\displaystyle \sum {x}={{\bar {x}}\times {n}}}$

${\displaystyle \sum {x}={12\times {10}}}$

${\displaystyle \sum {x}=120}$
σ=${\displaystyle {\sqrt {{\frac {\sum {x^{2}}}{n}}-({{\frac {\sum x}{n}})^{2}}}}}$
σ=${\displaystyle {\sqrt {{\frac {1530}{10}}-{144}}}}$
σ=${\displaystyle {\sqrt {{\frac {1530}{10}}-{144}}}}$
σ=${\displaystyle {\sqrt {153-144}}}$
σ=${\displaystyle {\sqrt {9}}}$
σ=3

# Problem 2

Problem No.1 of excercise No.6.1
Find the Standard deviation for the following data.

 x 03 08 13 18 23 f 07 10 15 10 08

# Solution:

 x f fx x² fx² 03 07 021 009 0063 08 10 080 064 0640 13 15 195 169 2535 18 10 180 324 3240 23 08 184 529 4232 n=50 Σfx=660 Σfx²=10710

Standard deviation σ=${\displaystyle {\sqrt {{\frac {\sum {fx^{2}}}{n}}-({{\frac {\sum fx}{n}})^{2}}}}}$
σ=${\displaystyle {\sqrt {{\frac {10700}{50}}-({{\frac {660}{50}})^{2}}}}}$
σ=${\displaystyle {\sqrt {214-174.24}}}$
σ=${\displaystyle {\sqrt {39.96}}}$
σ=6.3

# Problem 3

Problem No.5 of excercise No.6.3
Find the varience and Standard deviation for the following data.

 class intervals(CI) 30-34 34-38 38-42 42-46 46-50 50-54 freequency(f) 04 07 09 11 06 03

# Solution:

 C.I. f x fx d=${\displaystyle {\frac {x-A}{c}}}$ fd d² fd² 30-34 4 32 128 -2 -8 4 16 34-38 7 36 252 -1 -7 1 7 38-42 9 40 360 0 0 0 0 42-46 11 44 484 1 11 1 11 46-50 6 48 288 2 12 4 24 50-54 3 52 156 3 9 9 27 n=40 Σfx=1668 Σfd=17 Σfd²=85

A=assumed average.
c=4
d=${\displaystyle {\frac {x-A}{c}}}$=${\displaystyle {\frac {32-40}{4}}}$=${\displaystyle {\frac {-8}{4}}=-2}$

assumed mean A=${\displaystyle {\frac {\sum fx}{n}}}$=${\displaystyle {\frac {1668}{40}}=41.7}$

Varience σ²=[${\displaystyle {\frac {\sum {fd^{2}}}{n}}-({{\frac {\sum fx}{n}})^{2}}]c^{2}}$

σ²=[${\displaystyle {\frac {85}{40}}-({{\frac {17}{40}})^{2}}]4^{2}}$

σ²=[2.125-0.180]16

σ²=[1.945]16

σ²=31.12
standard deviation, σ=${\displaystyle {\sqrt {varience}}}$
σ=${\displaystyle {\sqrt {31.12}}}$
σ=5.58