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## Statistics, Algebra, Math, Computer Programming Tutor

### Bellevue, NE (68005)

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# Matthew's Responses in WyzAnt Answers

#### Statistics question. I don't know how to answer the orange part.

Assume diastolic blood pressure (DBP) is normally distributed with mean μ = 80
mmhg and standard deviation s = 12 mmhg for 35-44 year old men. That is,
DBP ~ N(80, 144). Diastolic BP values greater than 100 are considered to be
severe hypertension, and DBP values in the range of 90 to 100 are called moderate
hypertension.
a. What proportion of men in this age group do we expect to find with severe
hypertension?
What proportion do we expect to find with moderate hypertension?
Suppose that a random sample of 300 men in this age group working in a certain
industry have their blood pressure checked. Twenty-two of the men are found to
have severe hypertension, and 58 men have moderate hypertension.
b. How do the observed number of men with moderate and severe hypertension
compare to the expected numbers?
What conclusions can you draw regarding the rates (proportions) of hypertension
among 35-44 year old men in this industry?
c. What are the mean and standard error of the sampling distribution for DBP for
samples of size n=300 from the population of 35-44 year old men?
Penelope S. Pekow Page 3 3/1/13
Individual Assignment 2_sp13.docx
d. The observed sample mean DBP of the 300 men from the selected industry was
x =81.3 mmhg.
Knowing that the sampling distribution is normal when the underlying population
distribution is normal, what is the probability of observing a sample mean of 81.3
mmhg or greater when n=300, the population mean μ=80 mmhg and population
variance s2=144 mmhg2?
What conclusions can you draw about diastolic blood pressure among men in
this industry based upon your results? Does this agree / disagree with your
answer to part (b)?

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Asked by Linderpal from Sacramento, CA
10

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

#### Statistic Question Binominal Distribution.

I don't know how to answer part (d).

2. A large national survey conducted in 1995 indicated that 18% of American adults

had ever been tested for HIV at some point in their life. Suppose that in 2012 we

take a simple random sample of 100 adults and find that 27 report that they have

ever been tested.

a. Assume that proportion of the population that has been tested for HIV has not

changed since 1995. What is the probability that 27 or more adults out of 100

have been tested?

b. What assumptions are you making about your data and the study design to

compute this probability? Clearly define the probability model that you are using.

c. Based on your model, what is the

( i) expected number tested for HIV out of a sample of 100?

(ii) the variance and standard deviation?

d. Based on this evidence, do you think that the proportion of adults tested for HIV

has changed since 1995? Why or why not?

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Asked by Linderpal from Sacramento, CA
11

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.

#### find the values of theta..... hardest trig Q..

please.. not too sure how to solve this question.. any idea....?

T=theta

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Asked by Tid from Alexandria, VA
10

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.

#### do your tutors train at home?

will tutors travel to the home ?

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Asked by Bill from Camarillo, CA
10

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.

#### I start college in may and i dont understand anythig about algebra

I graduated in 1988 i only learned adding subtracting and multiplication and division i start online classes in may and i dont understand anything about algebra

+ more- less
Asked by David from Lancaster, SC
00

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!

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