Question

The lifetime of a certain type of TV tube has a normal distribution with a mean of 57 and a standard deviation of 6months. What proportion of the tubes lasts between 58 and 60 months?Answer: %Note: round your answer to 2 decimal places

Answer 1

In general, the z-score formula states that

$\begin{gathered} Z=\frac{x-\mu}{\sigma} \\ \mu\rightarrow mean \\ \sigma\rightarrow standard\text{ deviation} \end{gathered}$

Thus, in our case, the two z-scores are

$\begin{gathered} Z_{58}=\frac{58-57}{6}=\frac{1}{6} \\ and \\ Z_{60}=\frac{60-57}{6}=\frac{3}{6}=\frac{1}{2} \end{gathered}$

Use a z-score table to find the values of P(x<58) and P(x<60), as shown below

$\begin{gathered} \Rightarrow P(x<58)=0.56618 \\ P(x<60)=0.69146 \end{gathered}$

Hence,

[tex]\Rightarrow P(58Thus, the rounded answer is 12.53%