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Statistics question

2019 · 10 Jan · Shift 2 · Q39
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Statistics question

2019 · 10 Jan · Shift 2 · Q39

JEE MainMathematicsStatisticsMCQ+4 / −1
If mean and standard deviation of 5 observations x1, x2, x3, x4, x5 are 10 and 3, respectively, then the variance of 6 observations x1, x2, ….., x5 and –50 is equal to
  1. A
    582.5
  2. B
    507.5
  3. C
    586.5
  4. D
    509.5
View written solutionFree

Correct answer: B

  1. Given data for 5 observations

Mean =10=10=10, so

x1+x2+x3+x4+x55=10\frac{x_1+x_2+x_3+x_4+x_5}{5}=105x1​+x2​+x3​+x4​+x5​​=10

Hence,

x1+x2+x3+x4+x5=50x_1+x_2+x_3+x_4+x_5=50x1​+x2​+x3​+x4​+x5​=50

Standard deviation =3=3=3, so variance =32=9=3^2=9=32=9.

Using variance formula for 555 observations:

σ2=∑xi25−(∑xi5)2\sigma^2=\frac{\sum x_i^2}{5}-\left(\frac{\sum x_i}{5}\right)^2σ2=5∑xi2​​−(5∑xi​​)2

Thus,

9=∑xi25−1029=\frac{\sum x_i^2}{5}-10^29=5∑xi2​​−102 9=∑xi25−1009=\frac{\sum x_i^2}{5}-1009=5∑xi2​​−100 ∑xi25=109\frac{\sum x_i^2}{5}=1095∑xi2​​=109 ∑xi2=545\sum x_i^2=545∑xi2​=545
  1. Now include the 6th observation −50-50−50

New sum:

50+(−50)=050+(-50)=050+(−50)=0

So new mean is

xˉ=06=0\bar{x}=\frac{0}{6}=0xˉ=60​=0

New sum of squares:

545+(−50)2=545+2500=3045545+(-50)^2=545+2500=3045545+(−50)2=545+2500=3045
  1. Find the new variance

Variance of 6 observations:

σ2=30456−(0)2\sigma^2=\frac{3045}{6}-(0)^2σ2=63045​−(0)2 σ2=507.5\sigma^2=507.5σ2=507.5
  1. Match with options

The required variance is

507.5\boxed{507.5}507.5​

which corresponds to Option B.

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