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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,

$\mathrm{Î\S }=$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,

$\stackrel{¯}{X}~N(0.88,0.176)$

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

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

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

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

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