Variance for y =

\( \dfrac{1}{n^2}[ N\displaystyle\sum y_i^2-(\displaystyle\sum y_i)^2] \)

= \( \dfrac{1}{10}\times 40=4\)

Standard deviation (σ_{2}) = \( \sqrt{Variance}\)

= \( \sqrt{4}=2\)

CV (shares X) =

\( \dfrac{σ_2}{y}\times 100\) = \( \dfrac{2}{105}\times 100\)

= 1.9 = 11.58

C.V. of prices of shares X is greater than the C.V. of prices of shares Y.

Thus, the prices of shares Y are more stable than the prices of shares X.

Answered by Pragya Singh | 2 years 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 |