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6) The length of a phone conversation is normally distributed with a mean of 8 minutes and a standard deviation of 0.6 minutes. What is the probability that a conversation lasts less than 9 minutes?

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We are given a normal distribution with a mean of 8 and a standard deviation of 0.6. Since the probability is normally distributed we need first to find the z-score of the distribution, to do that we use the following formula:


z=(X-\mu)/(\sigma)

We have the following values:


\begin{gathered} X=9 \\ \mu=8 \\ \sigma=0.6 \end{gathered}

Replacing the values we get:


z=(9-8)/(0.6)=1.67

Now we use a normal distribution table to determine the probability for this z-score. That probability is:


P=0.9522\approx95.22percent

Therefore, the probability is 95.22 %

User Johneth
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