Question

Show all of your work, if using a calculator or Excel – Show formula and variables used to solve problem.

The SMC technical services department has embarked on a quality improvement effort. Its first project relates to maintaining the target upload speed for its Internet service subscribers. Upload speeds are measured on a standard scale in which the target value is 1.0. Data is collected over the past year indicate that the upload speed is approximately normally distributed, with a mean of 1.005 and a standard deviation of 0.10. Each day, one upload speed is measured. The upload speed is considered acceptable if the measurement on the standard scale is between 0.95 and 1.05.

1. Assuming that the distribution of upload speed has not changed from what is was in the past year, what is the probability that the upload speed is. (Answers should be 4 decimal places)

c. between 1.0 and 1.05?

d. less than 0.95 and greater than 1.05?

2. The objective of the operations team is to reduce the probability that the upload speed is below 1.0. Should the team focus on process improvement that increases the mean upload speed to 1.05 or on a process improvement that reduces the standard deviation of the upload speed to 0.075? Explain you answer using statistics

Answer #1

for normal distribution z score =(X-μ)/σx | |

here mean= μ= | 1.005 |

std deviation =σ= | 0.100 |

c)

probability
=P(1<X<1.05)=P((1-1.005)/0.1)<Z<(1.05-1.005)/0.1)=P(-0.05<Z<0.45)=0.6736-0.4801=0.1935 |

d)

probability
=1-P(0.95<X<1.05)=1-P((0.95-1.005)/0.1)<Z<(1.05-1.005)/0.1)=1-P(-0.55<Z<0.45)=1-(0.6736-0.2912)=0.6176 |

2)

for 1st case with mean =1.0:

probability =P(X>1)=P(Z>(1-1.05)/0.1)=P(Z>-0.5)=1-P(Z<-0.5)=1-0.3085=0.6915 |

for 2nd case with standard deviation of 0.075:

probability =P(X>1)=P(Z>(1-1.005)/0.075)=P(Z>-0.07)=1-P(Z<-0.07)=1-0.4721=0.5279 |

since increasing mean will increase probability of upload speed above 1.0 , one should focus on increasing the mean mean upload speed to 1.05

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