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?

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

Step-by-step 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()

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

                                    Z  =    ~ 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 =    = -1

   For 59 minutes, z-score is =    = 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.

Who knows bruh i’m ngl don’t even know the answer i’m just doing this cuz it’s mandatory