76. The attention span of a two-year-old is exponentially distributed with a mean of about eight minutes. Suppose we randomly survey 60 two-year-olds.

a. In words, X=

b. $X~$

c. In words$\stackrel{-}{X}$=

d. $\stackrel{-}{X}~$

e. Before doing any calculations, which do you think will be higher? Explain why.

i. The probability that an individual attention span is less than ten minutes.

ii. The probability that the average attention span for the 60 children in less than ten minutes?

f. Calculate the probabilities in part e.

g. Explain why the distribution for $\stackrel{-}{X}$ is not exponential.

The mean is $8$minutes

The standard deviation is $8$

sample size n =$60$

X is defined as the attention span of a two-year-old

We know,

mean ${\mu }_x=8$which is exponentially distributed

now using,

$X~\mathrm{Exp}\left(\frac{1}{{\mu }_x}\right)$

On putting the values we get

$\Rightarrow X~\mathrm{Exp}\left(\frac{1}{8}\right)$

The distribution is$X~\mathrm{Exp}\left(\frac{1}{8}\right)$

$\stackrel{-}{X}$is defined as the mean of the average attention span of two-year-old

We know,

mean ${\mu }_x=8$which is exponentially distributed

The standard deviation ${\sigma }_x=8$

sample size n =$60$

now using,

$\overline{X}~N\left({\mu }_x,\frac{{\sigma }_x}{\sqrt{n}}\right)$as it is normally distributed

Putting the values, we get the distribution

$=\overline{X}~N\left(8,\frac{8}{\sqrt{60}}\right)$

The standard deviation is more modest, so there is more region under the normal curve. For example $60$when contrasted with an example of one (a solitary individual), the standard deviation will be less and the qualities will bunch all the more firmly around the mean, and larger qualities will be more uncommon.

Hence The probability that the average attention span for the $60$children in less than ten minutes will be higher.

Considering that $P(x<10)$be the probability that the individual attention span is less than ten minutes.

Now for an exponential distribution with parameter $𝜆$the probability is$P(x\le a)=e^{-\frac{a}{\lambda }}$

localid="1652452836040" $$\begin{gathered} =P(x<10)=e^{-\frac{10}{0.125}} \\ =0.713 \end{gathered}$$

The probability that an individual attention span is less than ten minutes is $0.713$.

$P(\stackrel{-}{X}<10)$is the probability that the average attention span for the 60 children is less than ten minutes.

Using a calculator we get,

localid="1652453012848" $$\begin{gathered} P(\stackrel{-}{X}<10)=\text{normalcd}(10,8,1.033) \\ =0.973 \end{gathered}$$

According to the central limit theorem, the distribution for $\stackrel{-}{X}$is not exponential as the mean follows a normal distribution as n gets bigger.