Finding probability with mean and standard dev.

I had a question in class today, ("class question"!) that I got wrong, and couldn't seem to get a clear explanation for. The question was as follows:

If levels of lead in hair samples from that time period follow a normal distribution with mean 93 and standard deviation 16, find the probability that a randomly selected person from this time period would have a lead level of 130.5 ppm or higher.

I was extremely stumped and honestly have no idea how to do it. Any help would be great! thanks!

Use the z table.

\(\displaystyle z=\frac{130.5-93}{16}\)

Find z, look it up in the z table and subtract from 1.

The reason we subtract from 1 is because it is asking for "or higher". The region to the right of 130.5, but the normal curve comes from

negative infinity to the right.

but the z-score is 0.4904, if you subtract that from 1 it would be 50.96%???? That doesn't add up. Wouldn't you subtract it from 0.5?

Perhaps your z table is set up differently. You have the one you have to add .5 to because to goes from the center.

I have the other I am used to using.

I get a z score of 2.34

This corresponds to .4904+.5=.9904

1-.9904=.0096

Or, as you said, subtract from .5

.5-.4904=.0096

I got the same answer! 0.96%! Thanks! Now how about this question: White flies are devastating California crops. An area infested with white flies is to be sprayed with a chemical which is known to be 98% effective for each application. Assume a sample of 1000 flies is checked.

a.) Use the normal distribution to find the approximate probability that no more than 986 of the flies are killed in one application.

b.) Use the normal distribution to find the approximate probability that between 973 and 993 (inclusive) of the flies are killed in one application.

Using the normal to approximate a binomial is obsolete these days because we have computers. That is my feeling on the matter.

\(\displaystyle 1-\sum_{k=986}^{1000}\binom{1000}{k}(.98)^{k}(.02)^{1000-k}=.8975\)

Anyway, if it must be done that way.

\(\displaystyle np={\mu}=1000(.98)\)

\(\displaystyle \sqrt{npq}=\sigma=\sqrt{1000(.98)(.02)}\)

Use the continuity correction, 985.5, when setting up the z formula.

The solution should be close to the above binomial solution.

part b:

the continuity correction would be 972.5 and 993.5

Subtract the two results.

The binomial is \(\displaystyle \sum_{k=973}^{993}\binom{1000}{k}(.98)^{k}(.02)^{1000-k}=.949\)

I'm not sure what I'm looking at here, where do I start?

could you explain the notation for that first part?? Im not really understanding what your telling me... Sorry for my stupidity!

I'm sorry, I thought you were familiar with the binomial probability. You can google and find all about it, but what I have for the first part is

the probability that no more than 986 flies are killed. It can also be done with the normal approximation.

The reason being, back in the pre-computer days if we had large values to work with in the binomial, we could use the normal to approximate instead of having to sum up a lot of huge numbers.

For part a, using the normal:

\(\displaystyle z=\frac{985.5-980}{4.427}=1.24\)

Finding this in the table gives us .8925. Close to the actual we get with the binomial of .8975.

Instead of summing from 1 to 986, I summed from 986 to 1000 and subtracted from 1.

Do the same thing with the second part but use \(\displaystyle z=\frac{972.5-980}{4.427}\) and \(\displaystyle z=\frac{993.5-980}{4.427}\), then subtract the probabilities because it is between the values.

where did you come up with the 4.427?

so it's a standard deviation of 4.427 and a mean of 980? How?

Because in the normal approx. \(\displaystyle {\mu}=np\) and \(\displaystyle {\sigma}=\sqrt{np(1-p)}\)

So the answer to A would be 0.3925 or 39.25% and the answer to B would be 0.0444 or about 4.4%?

part a: add .5 to your result.

part b should be .949, or reasonably close to that. I used the binomial instead of the normal approx.

I need some direction with this math assignment.

Armed Forces Credit Union would like to examine the required sample size needed to be able to estimate the mean dollars that each card holder will spend each month. It would like to be within plus or minus $10 of the true mean with a 98% confidence level. The standard deviation is thought to be $500. How many card holders should be sampled?

After you have determined how many card holders should be sampled, Armed Forces Credit Union comes back and says that it will cost $5 per sample, and it was only planning on spending $10,000 on the sample. In a memo to the Armed Forces Credit Union product development team, indicate how many card holders should be sampled to meet the original requirements of the sample. Explain the trade-offs that will occur when you lower the sample to $10,000 to meet the budget.

Assignment Guidelines:

· Indicate how many card holders should be sampled to meet the original sample requirements of plus or minus $10 of the true mean with a 98% confidence level.

· Determine the new standard deviation and confidence intervals using the budget limit of $10,000.

o Be sure to show all work for your calculations.

· Address the following in 1,250–2,000 words:

o Explain the trade-offs that will occur when you lower the sample size to satisfy the $10,000 budget limit.

· Add your calculations and your work to the Word document.

· Save the file, and submit it to your instructor.

Please submit your assignment.

I came up with this:


Desire Cruz This is what I came up with but I am probably way off.

twenty samples at 500=20,000 (budget)
take the mean and square it this is then the standard deviation. 98% converts to .98= confidence level which converts to .02 Am I on the right track so far?