Is this unusual? What is the probability that you would have a mean reading rate of 92.9 wpm or more?

We use the calculator’s normalcdf capability and enter the ‘starting value’ �̅ , (92.9), then infinity

because we want everything from that number to infinity, but because we can’t enter infinity into the

calculator we type a very large number, hence 10^99, then the mean, 90, and ��̅, 2.3570.

How did we find ��̅ ? ��̅ = �

√�

��̅ = �

√� = 2.3570

The probability is 0.1093 (rounded to four places).

f) There is a 5% chance the mean reading speed of a random sample of 18 second grade students will

exceed what value?

μ=90

We need to find this value of �̅.

This a sampling distribution that is approximately normally distributed. We just found that ��̅ is 90 (the

same as μ because it’s approximately normal, so μ = ��̅ ). And the standard deviation ��̅ is 2.3570

(because we take the standard deviation divided by the square root of the sample size, 18).

We know the area to the right of �̅ is 0.05 because it said we want a 5% probability that it “will exceed”

this value.

Unfortunately, this is going to involve a LOT of trial and error.

We know that we need to enter normalcdf(??,10^99, 90, 2.3570)

[because we enter the starting value (which we do not yet know), then infinity because we want

everything from that number to infinity, but because we can’t enter infinity into the calculator we type a

very large number, hence 10^99, then the mean, 90, and ��̅ , 2.3570]

So, what number do we choose? Well, we can start with an educated guess.

We learned in part (e) that there was a 0.1093 probability that a student would have a reading rate of

92.9 or more. So we know that 0.1093 is the area to the right of 92.9. We want to have an area to the

right of 0.05. So this number will be larger than 92.9.

We also know from the empirical rule that since this is approximately normally distributed, we will

generally not have values that are more than three standard deviations above the mean.

So we will not need to try any values larger than 92.9 + 3*(2.3570) = 99.971

Because we are relying so much on technology for these answers, the only way to answer this question

is to try and try again. Yes there are other ways, but they are even more time-consuming and

cumbersome! (i.e. many formulas that need to be rearranged, etc.)

So let’s randomly try the value 95 and see what we get.

This only has an area of 0.0169 to the right, so our guess is too large.

Try again! This time I’ll try 93.5.

This has an area of 0.0687 to the right which is too much! So our guess is too small!

Now I’ll try 94.

Getting closer!! This is a slightly smaller area than what we want, so this number is too large.

Here I try 93.8 (which gives me an area of 0.0535) and 93.7 (which gives me an area of 0.05823).

THANK GOODNESS WE ONLY HAVE TO GET THIS TO THE NEAREST TENTH!!!!!!!!!!!!!

So the area closer to 0.05 is given when we use the value 93.8 for �̅ .

You may not get there in this many guesses. You may have to try 10 or more numbers before you find

the right one. (It’s annoying – I know.)