Question

Songs on an iPod David's iPod has about $10000$songs. The distribution of the play times for these songs is heavily skewed to the right with a mean of $225$seconds and a standard deviation of$60$ seconds.

(a) Explain why you cannot safely calculate the probability that the mean play time $\overline{x}$is more than $4$minutes $\left(240\right)$seconds for an SRS of $10$songs.

(b) Suppose we take an SRS of $36$songs instead. Explain how the central limit theorem allows us to find the probability that the mean playtime is more than $240$ seconds. Then calculate this probability. Show your work.

Short answer

(a) Population is neither normally distributed nor sample size is above $30$

(b) The probability is$0.0668$

Step-by-step solution

Part (a) Step-1 Given Information

Given in the question that,

Population mean$\mu$$=225$

Population standard deviation$\sigma$$=60$

Sample size$\left(n\right)=10$

we have to find the probability that the mean play time $\overline{x}$is more than $4$minutes $\left(240\right)$seconds for an SRS of $10$songs.

Part (a) Step-2 Explanation 

To calculate the probability either the sample size must be above $30$or the population must be normally distributed. Here, neither the original population is normally distributed nor the sample size is above $30$Thus, the required probability cannot be calculated here.

Part (b) Step-1 Given Information 

Given in the question that we take an SRS of $36$songs instead we have to Explain how the central limit theorem allows us to find the probability that the mean playtime is more than $240$seconds and calculate probability.

Part (b) Step-2 Explanation 

The central limit theorem states that sampling distribution of sample mean follows the normal distribution if the sample size is at least $30$, without considering of the nature of the distribution.

The sample size is higher than $30$. Thus, the required probability can be calculated here.

The probability that mean play time of song is above $240$seconds is calculated as follows:

$P\left(\overline{X}>240\right)=P\left(\frac{{\displaystyle \frac{x-\mu }{\sigma }}}{\sqrt{n}}>\frac{{\displaystyle \frac{240-\mu }{\sigma }}}{\sqrt{n}}\right)$

$=P\left(Z>\frac{{\displaystyle \frac{240-255}{60}}}{\sqrt{36}}\right)$ $P\left(Z>\frac{{\displaystyle \frac{240-225}{60}}}{\sqrt{36}}\right)$

$=P(Z>1.50)\left(Froms\tan dardnormaltable\right)$$=P\left(Z>1.50\right)\left(Froms\tan dardnormaltable\right)$

$=0.0668$

Thus, the required probability is $0.0668$