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I know how to calculate question which are phrased like so

A study of data collected at a company manufacturing flashlight batteries shows that a batch of 8000 batteries have a mean life of 250 minutes with a standard deviation of 20 minutes. Assuming a Normal Distribution, estimate:

(i) How many batteries will fail before 220 minutes?

Answer = 534.4

But I can not figure questions phrased like this:

Support call times at a technical support center are Normally distributed with a mean time of 8 minutes and 45 seconds and a standard deviation of 1 minute and 5 seconds. On a particular day, a total of 500 calls are taken at the centre. How many of these calls are likely to last more than 10 minutes

I dont understand how to find the z-score in this question as its to do with time?

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2 Answers 2

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To calculate the z-score you have to standardize the random variable. The support call time is distributed as $T\sim\mathcal N\left(8.75, (1\frac1{12})^2 \right)$

Reasoning: $45$ seconds are $0.75$ minutes. And 5 seconds are $\frac1{12}$ minutes.

Therefore $Z=\frac{T-8.75}{1\frac1{12}}=\frac{T-8.75}{\frac{13}{12}}$. Then it is asked for

$$P(T> 10)=1-P(T\leq 10)=1-\Phi\left(\frac{10-8.75}{\frac{13}{12}}\right)=1-\Phi\left(\frac{\frac54}{\frac{13}{12}}\right)=1-\Phi\left(\frac{15}{13}\right)$$

This is the probability that one arbitrary call last more than 10 minutes.

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  • $\begingroup$ So you add the 1 Minuit with the.05 seconds? I do not understand you are squaring $\endgroup$
    – Sean
    Commented Jan 10, 2020 at 13:12
  • $\begingroup$ The squaring is because of the notation. A normal distributed variable x is distributed as $\mathcal N(\mu, \sigma^2)$. The variance is $\sigma^2$ and therefore the square root of it is the standard deviation $\sigma$. $\textbf{So you add the 1 Minuit with the.05 seconds?}$ Yes, that´s right. $\endgroup$ Commented Jan 10, 2020 at 17:39
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The Z score is how many standard deviations above the mean 10 minutes is. It is $5/4$ minutes more than the mean of eight minutes and forty five seconds, and the standard deviation is $13/12$ minutes, so the Z score is $15/13$.

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