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=(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)=(58-57)/(6)=(1)/(6) \\ and \\ Z_(60)=(60-57)/(6)=(3)/(6)=(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%