Question

(1 point) 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 1

12.4%

Explanation:

Given:

Mean (μ) = 57

Standard deviation (σ) = 6

We can determine the percentage as follows:

`\begin{gathered} P(58\leq x\leq60)=\left((60-\mu)/(\sigma)\right)-\left((58-\mu)/(\sigma)\right) \\ \\ \text{ We replacing:} \\ \\ P\left(58\le\:x\le60\right)=P\left((60-57)/(6)\right)-P\left((58-57)/(6)\right) \\ \\ P\left(58\le\:x\le60\right)=P\left((3)/(6)\right)-P\left((1)/(6)\right) \\ \\ P\left(58\le\:x\le60\right)=P\left(0.5\right)-P\left(0.17\right) \\ \end{gathered}`

Determine these values with the help of the normal table, like this:

`\begin{gathered} P\left(58\le\:x\le60\right)=0.6915-0.5675 \\ \\ P\left(58\le\:x\le60\right)=0.124=12.4\% \end{gathered}`