The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases:

**(i) **If wrong item is omitted.

**(ii)** If it is replaced by 12

Asked by Abhisek | 1 year ago | 142

**(i)** If wrong item is omitted,

From the question it is given that,

The number of observations i.e. n = 20

The incorrect mean = 20

The incorrect standard deviation = 2

\(\overline {X}= \dfrac{1}{n} \displaystyle\sum_{i=1}^{20} (X_1) \)

\(10= \dfrac{1}{20} \displaystyle\sum_{i=1}^{20} (X_1) \)

\( \displaystyle\sum_{i=1}^{20} (X_1)=200\)

That is, incorrect sum of observations = 200

Correct sum of observations = 200 – 8 = 192

Correct mean = \( \dfrac{Correct\;sum}{19}\)

= \( \dfrac{192}{19}\) = 10.1

Standard deviation σ=

\( (σ)= \sqrt{\dfrac{1}{n} \displaystyle\sum_{i=1}^{n} X_1-\dfrac{1}{n^2} (\displaystyle\sum_{i=1}^{n}X)^2}\)

\( 2= \sqrt{\dfrac{1}{n} \displaystyle\sum_{i=1}^{n} (X_1^2-\overline X)^2}\)

\(4=\dfrac{1}{20} Incorrect \displaystyle\sum_{i=1}^{n}X_1^2 -100\)

\( Incorrect \displaystyle\sum_{i=1}^{n}X_1^2 =2080\)

Therefore,correct

\(correct \displaystyle\sum_{i=1}^{n}X_1^2= Incorrect \displaystyle\sum_{i=1}^{n}X_1^2 -(8)^2\)

= 2080 - 64

= 2016

Correct standard deviation =

\( \sqrt{\dfrac{correct \displaystyle\sum X_1^2}{n}-(correct\;mean)^2}\)

\( \sqrt{\dfrac{2016}{19}-(10.1)^2}\)

= \( \sqrt{1061.1- 102.1}\)

= \(\sqrt{4.09}\)

= 2.02

**(ii)** When 8 is replaced by 12,

Incorrect sum of observations = 200

Correct sum of observations = 200 – 8 + 12 = 204

Correct mean = \( \dfrac{Correct\;sum}{20}\)

= \( \dfrac{204}{20}=10.2\)

Standard deviation \( (σ)= \sqrt{\dfrac{1}{n} \displaystyle\sum_{i=1}^{n} X_1-\dfrac{1}{n^2} (\displaystyle\sum_{i=1}^{n}X)^2}\)

\( 2= \sqrt{\dfrac{1}{n} \displaystyle\sum_{i=1}^{n} (X_1^2-\overline X)^2}\)

\(4=\dfrac{1}{20} Incorrect \displaystyle\sum_{i=1}^{n}X_1^2 -100\)

\( Incorrect \displaystyle\sum_{i=1}^{n}X_1^2 =2080\)

Therefore,correct

\(correct \displaystyle\sum_{i=1}^{n}X_1^2= Incorrect \displaystyle\sum_{i=1}^{n}X_1^2 -(8)^2\)

= 2080 - 64

= 2160

Correct standard deviation=

\(\sqrt{\dfrac{correct \displaystyle\sum X_1^2}{n}-(correct\;mean)^2}\)

= \( \sqrt{\dfrac{2160}{20}}-(10.2)^2\)

= \( \sqrt{108-104.04}\)

= \( \sqrt{3.96}\)

= 1.98

Answered by Pragya Singh | 1 year agoFind the mean deviation from the mean and from a median of the following distribution:

Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |

No. of students | 5 | 8 | 15 | 16 | 6 |

The age distribution of 100 life-insurance policy holders is as follows

Age (on nearest birthday) | 17-19.5 | 20-25.5 | 26-35.5 | 36-40.5 | 41-50.5 | 51-55.5 | 56-60.5 | 61-70.5 |

No. of persons | 5 | 16 | 12 | 26 | 14 | 12 | 6 | 5 |

Compute mean deviation from mean of the following distribution:

Marks | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 | 80-90 |

No. of students | 8 | 10 | 15 | 25 | 20 | 18 | 9 | 5 |

Find the mean deviation from the mean for the following data:

Classes | 95-105 | 105-115 | 115-125 | 125-135 | 135-145 | 145-155 |

Frequencies | 9 | 13 | 16 | 26 | 30 | 12 |

Find the mean deviation from the mean for the following data:

Classes | 0-100 | 100-200 | 200-300 | 300-400 | 400-500 | 500-600 | 600-700 | 700-800 |

Frequencies | 4 | 8 | 9 | 10 | 7 | 5 | 4 | 3 |