Previously, De Anza statistics students estimated that the amount of change daytime statistics students carry is exponentially distributed with a mean of $\backslash (0.88$. Suppose that we randomly pick $25$daytime statistics students.

a. In words,${\rm X}=\_\_\_\_\_\_\_\_\_\_\_\_$

b.${\rm X}~\_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)$

c.role="math" localid="1651578876947" $Inwords,\overline{X}=\_\_\_\_\_\_\_\_\_\_\_\_$

d. $\overline{X}~\_\_\_\_\_\_(\_\_\_\_\_\_,\_\_\_\_\_\_)$

e. Find the probability that an individual had between $\backslash )0.80and\backslash (1.00$. Graph the situation, and shade in the area to be determined.

f. Find the probability that the average of the 25 students was between $\backslash )0.80and\$1.00$. Graph the situation, and shade in the area to be determined.

g. Explain why there is a difference in part e and part f.

the amount of change daytime statistics students carry is exponentially distributed with a mean of $\$0.88$. Suppose that we randomly pick $25$daytime statistics students.

According to the given information, the random variable is defined as,

${\rm X}=$amount of change students carry

According to the given information, the random variable can be expressed as,

$X~E(0.88,0.88)$

The other words of definition as follows:

$\overline{X}=$average amount of change carried by a sample of $25$ students.

In other words, part d can be expressed as,

$\overline{X}~N(0.88,0.176)$

the probability that an individual had between $\$0.80and\$1.00.$

$$\begin{gathered} P(0.8<x<1)=(1-e^{-0.88\times 1})-(1-e^{-0.88\times 0.8}) \\ P(0.8<x<1)=0.4946-0.4127 \\ P(0.8<x<1)=0.0819 \end{gathered}$$

the probability that the average of the $25$students was between $\$0.80and\$1.00.$

$P\left(\frac{(0.8-0.88)}{0.88/\sqrt{(}25)}\le Z\le \frac{(1-0.88)}{0.88/\sqrt{\left(25\right)}}\right)=P(-0.4545\le Z\le 0.68181)=0.1882$

The distributions are different. Part a is exponential and part b is normal.