Age of 6 members in a family are 17 years, 36 years, 12 years, 16 years, 63 Years and 39 years. Wha

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05 Jul 2017 Standard Deviation

Age of 6 members in a family are 17 years, 36 years, 12 years, 16 years, 63 Years and 39 years.
What is the population standard deviation of their ages?

OPtion

1) 17.76
2) 15.23
3) 18.37
4) 16.69
5) 17.67
6) 16.96
7) 15.32
8) 18.03
9) 17.86
10)None of these

Solution

Mean of 6 ages = (12+16+17+36+39+63)/6 = 30.5
Sum of squared difference from mean for each number = (30.5-12)^2 + (30.5-16)^2 + (30.5-17)^2 + (30.5-36)^2 + (30.5-39)^2 + (30.5-63)^2
=1893.5

Average of Squared difference (Variance) = 1893.5/6 = 315.58
Population Standard Deviation = Square root of variance = 315.58^1/2 = 17.76

Correct Option: 1 Best Solution (11)

Mayur Shinde   7 years AGO

1) Calculate Mean of giving data

=> 17+36+12+16+63+39/6
=> 183/6
=> 30.5

2)Subtract mean form each data point and find square of diff.

=>17-30.3 = (-13.5)*(-13.5) = 182.25
=> 36-30.5 = 5.5*5.5 = 30.25
=> 12-30.5 =(-18.5)*(-18.5)= 342.25
=> 16-30.5 = (-14.5)*(-14.4) = 210.25
=> 63-30.5 = 32.5*32.5 = 1056.25
=> 39-30.5 = 8.5*8.5 = 72.25

3) Calculate te mean of the squared differences

=>182.25+30.25+342.25+210.25+1056.25+72.25/6

=> 1893.75/6

=> 315.62

so variance is 315.62

4) population standard deviation is square root of the variance

=> √315.62
=> 17.76

So option 1 is correct one

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G.S.Kantharao   7 years AGO

Ages are 17,36,12,16,63,39
Average=30.5
Deviations=(30.5-17)^2+(30.5-36)^2+(30.5-12)^2+(30.5-16)^2+(30.5-63)^2+(30.5-39)^2=(13.5)^2+(-5.5)^2+(18.5)^2+(14.5)^2+(-32.5)^2+(-8.5)^2=182.25+30.25+342.25+210.25+1056.25+72.25=1893.5
Average=1893.5/6 = 315.583333...
Population Standard deviation = (315.583333....)^(1/2)=17.76.
Hence the correct option is 1👍

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Mukund Manohar   7 years AGO

Mean of 6 ages = (17+36+12+16+63+39)/6 = 30.5
Sum of squared difference from mean of each number = (30.5-17)^2 + (30.5-36)^2 + (30.5- 12)^2 + (30.5-16)^2 + (30.5-63)^2 + (30.5-39)^2
= 182.25 + 30.25 + 342.25 + 210.25 + 1056.25 + 72.25
= 1893.5
Average of Squared Difference(Variance) = 1893.5/6 = 315.58
Standard Deviation = Square Root of Variance = 315.58^(1/2) = 17.76

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SAYAN CHATTERJEE   7 years AGO

avg.= 1/6*(17+36+12+16+63+39)=30.5
sd.= root(1/6*((30.5-17)^2+(30.5-36)^2+(30.5-12)^2+(30.5-16)^2+(30.5-63)^2+(30.5-39)^2)
= 17.76
Option 1.

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ganesh chemanchula   7 years AGO

First find the mean of all ages.
Then subtract each age from mean value and square it.
Then find the mean of the sum of the squares
find the root of the value which is population standard deviation.
According to the problem, the answer will be 17.76. Option 1.

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Ashish Chaurasia   7 years AGO

first calculate mean of a given number:-
=(17+36+12+16+63+39)/6
=30.5
then subtract the from the given number and square it:-

= (17-30.5)^2+(36-30.5)^2+(12-30.5)^2+(16-30.5)^2+ (63-30.5)^2+(39-30.5)^2
=1893.5

then divide the number by number of members in a family=(1893.5)/6
= 315.583
then find square root of this number :-
= (315.583)^1/2
=17.76

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Maxx   7 years AGO

First calculate mean of 6 people=30.5
Calculate square of difference with mean of each person age individually and take the sum of these=1893.5
Divide 1893.5/6=315.583333=x
Take sqrt(x)=17.764

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Tanya Mahajan   7 years AGO

variance is 315.5833...
sqr.root of variance is 17.76 aprox

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somaiah tummala   7 years AGO

before finding the S.D we have to find mean and variance values

mean=17+36+12+16+63+39/6
=183/6
=30.5
now find variance

variance=(17-30.5)^2+(36-30.5)^2+(12-30.5)^2+(16-30.5)^2+(63-30.5)^2+(39-30.5)^2/6
=182.25+30.25+342.25+210.25+1056.25+72.25/6
=1893.5/6
=315.5833

now standard deviation = square root (315.5833)
=17.76

so option 1 is correct

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NABA KISHORE PADHY   7 years AGO

Mean age of 6 members = (17+36+12+16+63+39)/6=183/6=30.5
Answer is [{(17-30.5)^2+(36-30.5)^2+(12-30.5)^2+(16-30.5)^2+(63-30.5)^2+(39-30.5)^2}/6]^(1/2)
=315.58^(1/2)=17.76 approximately

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S.MUKESH KANNA   7 years AGO

Population standard deviation=sqrt of summation of(each value -mean value )power2divided by total number of values

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