I have no idea how to do these questions,I only know parts a(i) and b(i).
Also,are a(iii) and b(ii) conditional probability questions?
Find the probability (percentage) of staff rated as poor.
Take 1 - P(poor)
This gives you the sum percentages of staff with fair and good ratings.
Then, divide this by 2 to get the percentage of fair and good ratings.
So, now depending on what type of z table you have, you can proceed with two different ways.
Find the z value for P(X > z) = P(Good)
or Find the z value for P(X < z) = P(Poor) + P(Fair)
When you got the z value, convert it into X.
a(ii) Solving a(i) you know that the probability for a salesman to be rated poor equals 0.1288. Hence, 1-0.1288 of chance is left to be split among rates "fair" and "good". Meaning that there is an equal chance for a salesman to be rated "fair" and "good", each rate belonging the same probability (1-0.1288)/2=0.4356. So the surface under the probability density function is split into three parts: the first (leftmost) one measures 0.1288, and middle and the rightmost part each measuring 0.4356. All you have to do is find the value a having the property that 1-P(X<a)=0.4356. Can you do that?
For the second one, you know that:
This means, on average, 490 persons get the SMS within 3.5 seconds with a standard deviation of 9.8 people. If there are a total of 500 persons, more than 12 get the SMS after 3.5 seconds, that means there less than 488 people get the SMS within 3.5 s.
I think you can complete this now
P(Poor) = 0.1288
P(Fair) + P(Good) = 1- 0.1288 = 0.8712
P(Good) = 0.8712/2 = 0.4356
P(X > z) = 0.4356
P(X < z) = 1-0.4356 = 0.5644
The z value corresponding to this probability is 0.162 from my z table.
Then, I use:
Hence, a staff needs at least a score of 50.85 to be classified as 'Good'.
My z table gives the probabilities of z values to the left of the z value. I cannot read the probability of the z value given the probability after that value.
That is, I cannot find the z value for P(X > z), but I can find for P(X < z).
However, I could take the value directly, but this will require that I also invert the sign of the z value later. To avoid confusion, I already take the precautions beforehand.
The third part is some sort of conditional probability.