A set of final exam marks in a stats course is normally distributed, with a mean of 73 and a standard deviation of 9.
What is the probability that a student scored below 85 on this exam? Round to 4 decimal places as needed.
I don't understand this question either. Any help would be much appreciated ~~"..
Take a look at
This normal graph is brilliantly interactive. You slide the z value about and it gives you the probability as a percentage.
For your question you need to choose the display option 'up to z'
Now to decide what z is for your question.
Clearly, you couldn't have a separate graph for every possible mean and every possible standard deviation. You'd need an infinite number of them!!!
So statisticians have just one, the standard normal graph, which has a mean of zero and a sd. of 1.
So you have to convert your question into the standard graph version.
The score is 85 so it is 85 - 73 from the mean.
So imagine the whole graph shifted 73 places left. Now the mean is at zero and your 85 has become 12
So the equivalent question would be a mean of zero and a score of 12.
But the sd. is also wrong. The graph needs rescaling so that it has a sd. of 1.
You do that by dividing by 9.
Thus mean 73, sd 9 and score 85 becomes mean 0, sd. 1 and score
On the graph move the sllder until you get this z value and there's the probability.
In an exam they won't let you carry a computer in and go on-line, so how do you do the question then?
Look below the graph and you'll find a table of probabilities for given z values.
This table starts at the half way value of z = 0 and goes up to a z value of 3.09. It saves space by not giving the other half of the table, which would be the same by symmetry.
So you look for your z value, read off the probability, and add 0.5 for the missing half of the table.
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei