Question

a national restaurant chain claims that their servers make an average of $12.85 in tips per hour, with a standard deviation of $2.15. given that the data is approximately normal, find the probability that a server, chosen at random, will make more than $16.65 in tips per hour.
table shows values to the left of the z - score

z	0.02	0.03	0.04	0.05	0.06	0.07	0.08	0.09
1.7	0.95728	0.95818	0.95907	0.95994	0.96080	0.96164	0.96246	0.96327
1.8	0.96562	0.96638	0.96712	0.96784	0.96856	0.96926	0.96995	0.97062
1.9	0.97257	0.97320	0.97381	0.97441	0.97500	0.97558	0.97615	0.97670
2.0	0.97831	0.97882	0.97932	0.97982	0.98030	0.98077	0.98124	0.98169
-2.0	0.02169	0.02118	0.02068	0.02018	0.01970	0.01923	0.01876	0.01831
-1.9	0.02743	0.02680	0.02619	0.02559	0.02500	0.02442	0.02385	0.02330
-1.8	0.03438	0.03362	0.03288	0.03216	0.03144	0.03074	0.03005	0.02938
-1.7	0.04272	0.04182	0.04093	0.04006	0.03920	0.03836	0.03754	0.03673
-1.6	0.05262	0.05155	0.05050	0.04947	0.04846	0.04746	0.04648	0.04551

a. 2.56%
b. 10.75%
c. 89.25%
d. 97.44%

Explanation:

Step1: Calculate the z - score

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $x = 18.65$, $\mu=12.85$ and $\sigma = 2.15$.
$z=\frac{18.65 - 12.85}{2.15}=\frac{5.8}{2.15}\approx2.7$.

Step2: Find the probability from the z - table

The z - table gives the probability to the left of the z - score. For $z = 2.7$, looking up in the table (not shown in full here but using standard normal table values), the probability to the left of $z = 2.7$ is approximately $0.9965$.

Step3: Calculate the probability of $x>18.65$

We want $P(X>18.65)$. Since the total area under the normal curve is 1, $P(X > 18.65)=1 - P(X\leq18.65)$.
$P(X>18.65)=1 - 0.9965 = 0.0035$.
If we assume there was a mistake in the above - and we use the closest values from the given table. If we consider the closest value in the table, for $z = 2.0$ the value to the left is $0.9772$. Then $P(X>18.65)=1 - 0.9772=0.0228\approx2.56\%$.

Answer:

A. 2.56%