#4. Two employees are chosen at random (employees are male/female, manager/teacher). Calculate P[male teacher and any female].
There's a simple-minded way to do this, which I usually use first. P[male teacher and any female] = (# of ways to choose a male teacher and any female) / (# of ways to pick two employees).
To find the # of ways to ..., use the combination formula (since order of selection doesn't matter). Note that there are a total of 86 male teachers + females, and a total of 100 employees.
Another way to do this, which can either serve as a check on your answer, or maybe you prefer to think it through in this manner: selections that would fail involve picking at least 1 male manager. You could think of the selections as occurring in two steps--select one person, then select another. So, selections that fail would be ones where you either (a) select a male manager and then anyone else, or (b) select anyone besides a male manager and then a male manager. Note the way that (b) is worded: you don't want to double-count selections that were counted in (a)
#9. 250 vehicles on a motorway. Estimate the # of vehicles travelling between 52 and 63 mph.
I have the feeling that they don't want you to count the # of squares there are in the height of each box (boxes for 40-50 mph, 50-60 mph, 60-70 mph, 70-75 mph, 75-80 mph, 80-100 mph). Why? Because they don't give a handy vertical scale/markings on the vertical axis. Morever, the range they question you about does not fall exactly on the ranges for the boxes.
So if I had to give an estimate, I would assume that the speeds are normally distributed. Does this sound okay?
Since they say that 55 vehicles were travelling over 75 mph, I would translate that to: P(X > 75) = 55/250 = .22. In a normal table, the probability .22 corresponds to a z = 0.77. So I'd say that (75 - mean) / stdev = 0.77. Now here's a problem, that there are two unknowns for this one equation. The best that I can come up with is to say that mean=70 (because it falls in the middle of the range of speeds, and the peak of the histogram occurs around there). This leads to stdev = 6.5.
To finish up, you would standardize X (the thing where you subtract the mean from each number then divide by stdev), and get a probability by looking up a normal table. Multiply probability by 250 to get the number the question is asking for.
Other comments are welcome.