Question

The playing time X of jazz CDs has the normal distribution with mean 52 and standard deviation 7; N(52, 7).

According to the 68-95-99.7 rule, what percentage of jazz CDs play between 38 and 59 minutes?

Answer 1

68% of jazz CDs play between 45 and 59 minutes.

Explanation:

The correct question is: The playing time X of jazz CDs has the normal distribution with mean 52 and standard deviation 7; N(52, 7).

According to the 68-95-99.7 rule, what percentage of jazz CDs play between 45 and 59 minutes?

Let X = playing time of jazz CDs

SO, X ~ Normal(
`\mu=52, \sigma^(2) =7`)

The z-score probability distribution for the normal distribution is given by;

Z =
`(X-\mu)/(\sigma)` ~ N(0,1)

Now, according to the 68-95-99.7 rule, it is stated that;

• 68% of the data values lie within one standard deviation points from the mean.

• 95% of the data values lie within two standard deviation points from the mean.

• 99.7% of the data values lie within three standard deviation points from the mean.

Here, we have to find the percentage of jazz CDs play between 45 and 59 minutes;

For 45 minutes, z-score is =
`(45-52)/(7)` = -1

For 59 minutes, z-score is =
`(59-52)/(7)` = 1

This means that our data values lie within 1 standard deviation points, so it is stated that 68% of jazz CDs play between 45 and 59 minutes.