## Answers

N = 20

Let x be the load capacities

x | |

11.0 | 65.61 |

20.0 | 0.81 |

24.0 | 24.01 |

13.0 | 37.21 |

16.0 | 9.61 |

24.0 | 24.01 |

18.0 | 1.21 |

7.0 | 146.41 |

20.0 | 0.81 |

22.0 | 8.41 |

17.0 | 4.41 |

18.0 | 1.21 |

28.0 | 79.21 |

27.0 | 62.41 |

20.0 | 0.81 |

15.0 | 16.81 |

12.0 | 50.41 |

11.0 | 65.61 |

31.0 | 141.61 |

28.0 | 79.21 |

Q.1.

a)

Mean of the population:

**Mean of the population is 19.1**

Standard deviation of the population:

(Round to 3 decimal)

**Standard deviation of the population is 6.402**

Coefficient of variation (CV) :

CV = 33.52%

**Coefficient of variation is 33.52%**

Q.2

b)

n = sample size = 9

Consider first 9 values from above 20 values as a sample.

x | |

11.0 | 36 |

20.0 | 9 |

24.0 | 49 |

13.0 | 16 |

16.0 | 1 |

24.0 | 49 |

18.0 | 1 |

7.0 | 100 |

20.0 | 9 |

Point estimate of mean:

**Point estimate of mean is 17**

Point estimate of Standard deviation:

(Round to 3 decimal)

**Point estimate of Standard deviation is 5.809**

Point estimate of Coefficient of variation (CV) :

CV = 34.17%

**Point estimate of Coefficient of variation is 34.17%**

c)

Standard error of the mean:

SE = 1.936 (Round to 3 decimal)

**Standard error of the mean is 1.936**

d)

90% confidence interval :

COnfidence level = c = 0.90

Sample size = n = 9

Sample mean = = 17

Population standard deviation = = 6.402

Here Population standard deviation is known so we use z interval.

90% confidence interval is

where zc is z critical value for (1+c)/2 = (1+0.90)/2 = 0.95

zc = 1.645 (From statistical table of z values, average of 1.64 and 1.65,(1.64+1.65)/2 = 1.645)

(Round to 3 decimal)

**90% confidence interval is (13.490, 20.510)**