Let the remaining two observations be x and y.

Therefore, the observations are 6, 7, 10, 12, 12, 13, x, y.

\( Mean,\overline X\)

\( =\dfrac{6+ 7 +10 +12 +12 +13+x+y}{8}=9\)

60 + x + y= 72

x + y = 12 ...........(1)

Variance = \( \dfrac{1}{n} \displaystyle\sum_{i=1}^{n} f_i (x_i-\overline {x} )^2 \)

9.25 = \( \dfrac{1}{8}[(-3)^2 + (-2)^2+1^2+3^2+4^2+x^2+y^2\)

\(-18(x+y)+2\times (9)^2]\)

9.25 =\( \dfrac{1}{8}[9 +4+1+9+9+x^2+y^2\)

\( -18\times 12+162]\)

9.25 = \( \dfrac{1}{8}[48+x^2+y^2-216+162]\)

9.25 = \( \dfrac{1}{8}[x^2+y^2-6]\)

x^{2 }+ y^{2}=80 ................(2)

From (1), we obtain

x^{2}+y^{2 }+ 2xy=144 ................(3)

From (2) and (3), we obtain

2xy = 64 ….....................(4)

Subtracting (4) from (2), we obtain

x^{2 }+ y^{2 }-2xy = 80- 64 =16

x - y = ± 4 .................(5)

Therefore, from (1) and (5), we obtain

x = 8 and y = 4, when x – y = 4

x = 4 and y = 8, when x – y = – 4

Thus, the remaining observations are 4 and 8.

Answered by Pragya Singh | 11 months 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 |