Matthew's Responses in WyzAnt Answers

Your sample means are normally distributed as well!  The mean of samples will be 80, the same as the population mean, but the standard deviation of the samples will be different from the population standard deviation (12 in this case, as you're given variance = 144, and we know standard deviation is the square root of variance).

You are trying to find the probability of the mean greater than or equal to x=81.3.  We need to convert this number to a Z-score, which would normally be Z = ( x - mean) / stdev, except the standard deviation is not 12, but 12/sqrt(300) in this case, as we're referring to the standard deviation of the sample means, not of the original data, and the sample size helps determine the standard deviation of the means.

So Z = ( x - mean )/ (stdev/sqrt(n)) = (81.3-80.0)/(12/sqrt(300)) = 1.876, which we round up to 1.88 for ease in looking up the probability on a standard normal distribution table.  Assuming your table lists probabiliities of Z less than given numbers, we need to look up the P(Z < 1.88) and subtract that quantity from 1 to get the right answer.

P(Z > 1.88) = 1 - P(Z < 1.88) = 1 - 0.9699 = 0.0301

Part d. is asking you to consider your result in part b. If the probability of the event is very low, perhaps more people are being tested. It's really hard to put too much stock into the results for 100 people... a "large national survey" suggests many thousands of people were asked. Small samples will vary more than a much larger sample (recall the relationship of standard deviation of sample means, it involves division by the square root of n... larger n, less standard deviation). So while there may be evidence from your small sample, it's not enough information to be confident that the population proportion has changed.

Let's use the method you listed above, but we need to correct the formula: (and note "T+a") is not the a that is in the first term, it is an angle, which I will represent with delta, δ):

a cos θ + b sin θ = c cos ( θ - δ )

The original equation, if we switch the order of the first two terms, will help us find a, b and c.

4 cos θ + 2 sin θ = sqrt(5)

From the above equation, a=4, b=2, and c=sqrt(a²+b²)=(sqrt(4²+2²))= sqrt(16+4)=sqrt(20).

Substituting these values into the method:

4 cos θ + 2 sin θ = sqrt(20) * cos ( θ - δ ) = sqrt(5)

The last equality, sqrt(5), comes from the original equation which looks just like the method on the left hand side of the equation.

Dividing both sides by sqrt(20):

cos ( θ - δ ) = sqrt(5) / sqrt(20) = sqrt(5/20) = sqrt(1/4) = 1/2.

So now we have cos ( θ - δ ) = 1/2.  Taking the inverse cosine on both sides:

θ - δ = cos^-1(0.5). 

Adding delta to both sides:

θ = cos^-1(0.5) + δ.

We need to determine δ to solve this equation. 

A handy relationship is that tan δ = b/a, so δ = tan^-1(b/a). 

b=2 and a=4, so δ = tan^-1(1/2) = tan^-1(0.5).

The phase shift δ in the cosine curve cos ( θ - δ ) is tan^-1(0.5), which is ~.464 radians or ~26.56 degrees.

So θ = cos^-1(0.5) + 26.56.  So for every angle where cos^-1(angle)=0.5, that is our solution (provided it lies within the 0 to 360 degree range stipulated at the beginning of the problem.

We know that this happens when the angle is -60 and +60 degrees, and that pair repeats every 360 degrees.  Adding -60 to 26.56 gives us an angle out of range.  60 works, as does -60+360=300.  60+360=420 is out of range too.  So 60 and 300 are valid angles, which result in theta values of:

θ = 60 + 26.56 = 86.56 degrees and θ = 300 + 26.56 = 326.56 degrees.

The curve will be a standard cosine curve shifted to the right 26.56 degrees and multiplied by sqrt(20), such that its range will not be +1 to -1 but +(sqrt(20)) to -(sqrt(20))... a 'taller' curve if you will by almost 4.5 times.

 

I let the client decide.  If the parent is comfortable with me coming to their house, I will.  My biggest concern is whether or not distractions at the home, such as siblings, electronic distractions, etc., will complicate the instruction.  I prefer "neutral ground" like a public library, given a choice, where it is quiet and distraction-free.

If you'll be using a textbook for class, perhaps you can buy it in advance, or gain access to the online textbook so you can begin getting familiar with the concepts in algebra.  That might help lessen your anxiety.  Plus remember you're going into the class to learn; no one expects you to understand the concepts in advance.  I know after 25 years starting college might be stressful, but, as an instructor (and a class of 1988 grad too I might add), we're there to help.  Don't be afraid to speak up when there's something you don't understand.  Best of luck to you!