Answer:
We conclude that the sample is from a population of songs with a mean greater than 210 seconds.
Step-by-step explanation:
We are given that a simple random sample of 40 current hit songs results in a mean length of 252.5 seconds. 
Assume that the standard deviation of song lengths is 54.5 sec.
Let  = <u><em>population mean length of the songs</em></u>
 = <u><em>population mean length of the songs</em></u>
So, Null Hypothesis,  :
 :  
  210 seconds      {means that the sample is from a population of songs with a mean smaller than or equal to 210 seconds}
 210 seconds      {means that the sample is from a population of songs with a mean smaller than or equal to 210 seconds}
Alternate Hypothesis,  :
 :  > 210 seconds      {means that the sample is from a population of songs with a mean greater than 210 seconds}
 > 210 seconds      {means that the sample is from a population of songs with a mean greater than 210 seconds}
The test statistics that will be used here is <u>One-sample z-test statistics </u>because we know about population standard deviation;
                               T.S.  =   ~ N(0,1)
  ~ N(0,1)
where,  = sample mean length of songs = 252.5 seconds
 = sample mean length of songs = 252.5 seconds
              = population standard deviation = 54.5 seconds
 = population standard deviation = 54.5 seconds
             n = sample of current hit songs = 40
So, <u><em>the test statistics</em></u> =  
                                    =  4.932
The value of z-test statistics is 4.932.
Now, at 0.05 level of significance, the z table gives a critical value of 1.645 for the right-tailed test.
Since the value of our test statistics is more than the critical value of z as 4.932 > 1.645, so <u><em>we have sufficient evidence to reject our null hypothesis</em></u> as it will fall in the rejection region.
Therefore, we conclude that the sample is from a population of songs with a mean greater than 210 seconds.