# Consider Dataset A for answeri

Consider Dataset A for answering the questions that followsbelow. a. Calculate the measures of central tendencies for VariableX and Variable Y. i. Mean ii. Median iii. Mode iv. Midrange v. Whatcan you say about the skewness of X and Y variables? b. Calculatethe measures of variations for Variable X and Variable Y. i. Rangeii. Variance iii. Standard Deviation iv. Coefficient of Variationv. Which is more variable, X or Y? Why? c. Calculate the measuresof position for Variable X. i. Z-score of the mean value of X ii.Percentile rank of the maximum value of X iii. Check for anyoutliers in variable X

variable X:6,7,8,8,8,9,9,9,11,11

variableY:-2.77,-0.23,-0.29,-0.05,0.33,0.43,0.51,0.63,0.85,1.12

Answer:

Variable X:

————————————————————————–

X | (X – X̄)² |

6 | 6.76 |

7 | 2.56 |

8 | 0.36 |

8 | 0.36 |

8 | 0.36 |

9 | 0.16 |

9 | 0.160 |

9 | 0.160 |

11 | 5.760 |

11 | 5.760 |

X | (X – X̄)² | |

total sum | 86 | 22.40 |

n | 10 | 10 |

mean = ΣX/n = 86.000 / 10 = 8.6000 sample variance = Σ(X – X̄)²/(n-1)= 22.4000 / 9 = 2.489 sample std dev = √ [ Σ(X – X̄)²/(n-1)] = √ (22.4/9) = 1.5776

range=max-min = 11 – 6 = 5mid range=(max + min)/2= ( 11 + 6 ) /2 = 8.5

mode= highest frequency data = 8

coefficient of variation,CV=σ/µ= 0.183444

skewness using pearson coefficient of skewness,PC PC=3(mean-median)/std dev= 0.190160

IQR = Q3-Q1 = 1.75 1.5IQR = 2.625 lower bound=Q1-1.5IQR= 5.125 upper bound=Q3+1.5IQR= 12.125 outlier =values outside lower bound and upperbound total outlier below lower bound= 0total outlier above upper bound= 0**total outlier = 0**

**Z score of mean value = 0**

**Percentile Rank of maximum value = 1**

Variable Y

————————————-

X | (X – X̄)² | |

total sum | 0.53 | 10.71 |

n | 10 | 10 |

mean = ΣX/n = 0.530 / 10 = 0.0530 sample variance = Σ(X – X̄)²/(n-1)= 10.7120 / 9 = 1.190 sample std dev = √ [ Σ(X – X̄)²/(n-1)] = √ (10.712/9) = 1.0910

range=max-min = 1.12 – -2.77 = 3.89mid range=(max + min)/2= ( 1.12 + -2.77 ) /2 = -0.825Skewness= -2.1904

coefficient of variation,CV=σ/µ= 20.584407

lower bound=Q1-1.5IQR= -1.64 upper bound=Q3+1.5IQR= 2.08 outlier =values outside lower bound and upperbound total outlier below lower bound= 1total outlier above upper bound= 0**total outlier = 1**

**Thanks in advance!**

**revert back for doubt**

**Please upvote**