ASSIGNMENT 3

SUBMITTED BY :John Michael franklin

SECTION: A

ROLL NO:SMBA10080

Chapter 10

Q3)

Average Mean = x = (3.06 + 3.15 + 3.11 + 3.13 + 3.06 + 3.09) / 6 = 3.1

Range = R = (0.42 + 0.5 + 0.41 + 0.46 + 0.46 + 0.45 ) / 6 = 0.45

From table 10.3, for n=20, we have , A2 = 0.18, D3 = 0.41, D4 = 1.59

UCLx = x + A2 R

= 3.1 + (0.18 x 0.45)

= 3.181

LCLx= x – A2 R

= 3.1 – (0.18 x 0.45)

= 3.019

UCLR = D4 R

= 1.59 x 0.45

= 0.7155

LCLR = D3 R

= 0.41 x 0.45

= 0.1845

The largest sample mean = 3.15 , smallest = 3.06. Both are well within the control limits.

Similarly, the largest sample range is 0.50 and the smallest is 0.41.Both are well within the control limits.

Hence, the result suggests that the process is in control.

Q4) From the given data, then, using MS Excel, we compute as follows:

| Sample |

| 1 | 2 | 3 | 4 | 5 | 6 |

| 79.2 | 80.5 | 79.6 | 78.9 | 80.5 | 79.7 |

| 78.8 | 78.7 | 79.6 | 79.4 | 79.6 | 80.6 |

| 80 | 81 | 80.4 | 79.7 | 80.4 | 80.5 |

| 78.4 | 80.4 | 80.3 | 79.4 | 80.8 | 80 |

| 81 | 80.1 | 80.8 | 80.6 | 78.8 | 81.1 |

Total | 397.4 | 400.7 | 400.7 | 398 | 400.1 | 401.9 |

Mean | 79.48 | 80.14 | 80.14 | 79.6 | 80.02 | 80.38 |

Largest No. | 81 | 81 | 80.8 | 80.6 | 80.8 | 81.1 |

Smallest No. | 78.4 | 78.7 | 79.6 | 78.9 | 78.8 | 79.7 |

Range | 2.6 | 2.3 | 1.2 | 1.7 | 2 | 1.4 |

Then, using MS Excel, we compute as follow:

Average Mean = x | 79.96 |

Average Range = R | 1.866667 |

From table 10.3, for n=5, we have , A2 = 0.58, D3 = 0, D4 = 2.11

UCLx = x + A2 R

= 79.96+ (0.58 x 1.866667)

= 81.0446

LCLx= x – A2 R

= 79.96- (0.58 x 1.866667)

= 79.96 – 1.0846

=78.8754

UCLR = D4 R

= 2.11 x 1.866667

= 3.9457

LCLR = D3 R

= 0 x 1.866667

= 0

The largest sample mean is 80.38 and the smallest is 79.48. Both are well within the control limits.

Similarly, the largest sample range is 2.6 and the smallest...