Answer:
a.35.57%
b. 100.686
Step-by-step explanation:
To find the percentage of the test scores that exceeded 84, we need to standardize 84 as:
![z=\frac{84-m}{s} = \frac{84-80}{8.1}=0.37](https://tex.z-dn.net/?f=z%3D%5Cfrac%7B84-m%7D%7Bs%7D%20%3D%20%5Cfrac%7B84-80%7D%7B8.1%7D%3D0.37)
Where m is the mean of the scores and s is the standard deviation. Now, using the standard normal table, we can know the following probability:
P(Z > 0.37) = 0.3557
So, the percentage of the test scores that exceeded 84 were 35.57%
Now, to find the candidate's score that fell at the 98th percentile of the distribution we need to find th z-value that satisfies:
P(Z < z) = 0.98
So, using the standard normal table we have:
z = 2.06
Then, the candidate's score x is 100.686 and it is calculated as:
![z=\frac{x-84}{8.1}=2.06\\ x=2.06*8.1 + 84\\x=100.686](https://tex.z-dn.net/?f=z%3D%5Cfrac%7Bx-84%7D%7B8.1%7D%3D2.06%5C%5C%20x%3D2.06%2A8.1%20%2B%2084%5C%5Cx%3D100.686)