Question

In: Statistics and Probability

When a machine is properly calibrated, it takes an average time of 19.0 seconds to produce...

When a machine is properly calibrated, it takes an average time of 19.0 seconds to produce a widget, with a standard deviation of 2.8 seconds. For a simple random sample of n = 36 widgets, the sample mean production time is found to be x-bar = 20.2 seconds.

  1. If the machine was properly calibrated, what is the probability that the mean for this sample would be 20.2 seconds or more?

  1. Now that you know the probability of this sample mean occurring, what would you conclude about the calibration of the machine? Is it likely that it is calibrated properly?

Solutions

Expert Solution

Solution :

Given that,

mean = = 19.0

standard deviation = = 2.8

n = 36

= 19.0

= / n = 2.8 / 36 = 0.467

P( > 20.2 ) = 1 - P( < 20.2 )

= 1 - P[( - ) / < ( 20.2 - 19.0) / 0.467]

= 1 - P(z < 2.57)

Using z table,    

= 1 - 0.9949

= 0.0051

Probability = 0.0051

  

orchestra answered 2 years ago
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