****Can U Help Me Set These Up & What You Get For Comparison To Mine

sassytonigirl · 2016-01-31 12:48:39

I DO THE PROBLEMS OVER AND OVER AND THEN ITS LIKE I CANNOT FIGURE OUT HOW TO SET THEM UP AND THEN I QUESTION MY ANSWERS AND OF COARSE IF ONE LITTLE THING IS OFF THEN IT IS.....WRONG. CAN YOU HELP ME WITH THESE SO I CAN COMPARE WHAT I GET? THANKS!!!

4. A survey found that women's heights are normally distributed with mean 62.4in. and standard deviation 2.5 in. The survey also found that men's heights are normally distributed with a mean 69.1 in. and standard deviation 2.9. Complete parts a through c below.

a. Most of the live characters at an amusement park have height requirements with a minimum of 4 ft 9 in. and a maximum of 6 ft 3 in. Find the percentage of women meeting the height requirement.
The percentage of women who meet the height requirement is ________%.
(Round to two decimal places as needed.)

b. Find the percentage of men meeting the height requirement.
The percentage of men who meet the height requirement is _________%.
(Round to two decimal places as needed.)

c. If the height requirements are changed to exclude only the tallest 5% of men and the shortest 5% of women, what are the new height requirements?
The new height requirements are at least ________ in. and at most __________ in.
(Round to one decimal place as needed.)

6. The population of current statistics students has ages with mean μ and standard deviation σ. Samples of statistics students are randomly selected so that there are exactly 53 students in each sample. For each sample, the mean age is computed. What does the central limit theorem tell us about the distribution of those mean ages?
Choose the correct answer below.

A. Because n>30, the sampling distribution of the mean ages can be approximated by a normal distribution with mean μ and standard deviation σ divided by square root of 53.

B. Because n>30, the sampling distribution of the mean ages is precisely a normal distribution with mean μ and standard deviation σ/(divided by) square root of 53.

C. Because n>30, the sampling distribution of the mean ages can be approximated by a normal distribution with mean μ and standard deviation σ.

D.Because n>30, the central limit theorem does not apply in this situation.

7. Assume that women's heights are normally distributed with a mean given by μ=62.4 in, and a standard deviation given by σ=1.9 in.

(a) If 1 woman is randomly selected, find the probability that her height is less than 63 in.
(b) If 39 women are randomly selected, find the probability that they have a mean height less than 63 in.

(a) The probability is approximately _________.
(Round to four decimal places as needed.)

(b) The probability is approximately _________.
(Round to four decimal places as needed.)

8. A certain brand of candies have a mean weight of 0.8602 g and a standard deviation of 0.0511 A sample of these candies came from a package containing 464 candies, and the package label stated that the net weight is 396.4 g. (If every package has 464 candies, the mean weight of the candies must exceed 396.4/ 464 (Fraction)=0.8543 g for the net contents to weigh at least 396.4 g

a. If 1 candy is randomly selected, find the probability that it weighs more than 0.8543 g.
The probability is _________. (Round to four decimal places as needed.)

b. If 464 candies are randomly selected, find the probability that their mean weight is at least 0.8543 g.
The probability that a sample of 464 candies will have a mean of 0.8543 g or greater is _________.(Round to four decimal places as needed.)

c. Given these results, does it seem that the candy company is providing consumers with the amount claimed on the label?

▼ Yes or No_________because the probability of getting a sample mean of 0.8543 g or greater when 464 candies are selected

▼ is or is not_________ exceptionally small.

10. Use a normal approximation to find the probability of the indicated number of voters. In this case, assume that 144 eligible voters aged 18-24 are randomly selected. Suppose a previous study showed that among eligible voters aged 18-24, 22% of them voted.

Probability that fewer than 35 voted

The probability that fewer than 35 of 144 eligible voters voted is __________.(Round to four decimal places as needed.)

bobbym · 2016-01-31 13:29:57

Hi;

I am getting for the first,

4)

a) 98.46 %

b) 97.90 %

6) I like A.

sassytonigirl · 2016-01-31 14:14:39

What did you get for 4---- Letter c?

#7....this is what i got
a. 0.6255
b. 0.9756

#10...this is what i got
0.7157

Did you get the same thing for #7 and #10? and also let me know when you get the others above.

Thanks!

Tonnie

bobbym · 2016-01-31 17:15:32

Sorry, I did not even understand the question for 4c.

For 7a) I got an answer very close to that.

For 10) I got an answer which is close to yours.

sassytonigirl · 2016-01-31 23:28:29

Like so close that you think my answer is correct or you think I should rework number 7 and 10? What did you get for 7b?

what about this one?

8. A certain brand of candies have a mean weight of 0.8602 g and a standard deviation of 0.0511 A sample of these candies came from a package containing 464 candies, and the package label stated that the net weight is 396.4 g. (If every package has 464 candies, the mean weight of the candies must exceed 396.4/ 464 (Fraction)=0.8543 g for the net contents to weigh at least 396.4 g
a. If 1 candy is randomly selected, find the probability that it weighs more than 0.8543 g.
The probability is _________. (Round to four decimal places as needed.)
b. If 464 candies are randomly selected, find the probability that their mean weight is at least 0.8543 g.
The probability that a sample of 464 candies will have a mean of 0.8543 g or greater is _________.(Round to four decimal places as needed.)
c. Given these results, does it seem that the candy company is providing consumers with the amount claimed on the label?
▼ Yes or No_________because the probability of getting a sample mean of 0.8543 g or greater when 464 candies are selected
▼ is or is not_________ exceptionally small.

Thanks for your help

Bob · 2016-02-01 00:57:13

hi sassytonigirl

4c ?

They've given an 'x' and you've had to work out the probability. Now they want you to work out the reverse ... given P = 0.95 what x value does that correspond to ?

Bob

bobbym · 2016-02-01 04:06:01

Like so close that you think my answer is correct or you think I should rework number 7 and 10? What did you get for 7b?

My answers are a bit more accurate because I have better tables. Keep the answers you have.

I will work on 7b as soon as I can.

sassytonigirl · 2016-02-01 11:45:01

Bobbym,

I need this one more than 7b. I don't even know how to set this one up . Can you help?

8. A certain brand of candies have a mean weight of 0.8602 g and a standard deviation of 0.0511 A sample of these candies came from a package containing 464 candies, and the package label stated that the net weight is 396.4 g. (If every package has 464 candies, the mean weight of the candies must exceed 396.4/ 464 (Fraction)=0.8543 g for the net contents to weigh at least 396.4 g
a. If 1 candy is randomly selected, find the probability that it weighs more than 0.8543 g.
The probability is _________. (Round to four decimal places as needed.)
b. If 464 candies are randomly selected, find the probability that their mean weight is at least 0.8543 g.
The probability that a sample of 464 candies will have a mean of 0.8543 g or greater is _________.(Round to four decimal places as needed.)
c. Given these results, does it seem that the candy company is providing consumers with the amount claimed on the label?
▼ Yes or No_________because the probability of getting a sample mean of 0.8543 g or greater when 464 candies are selected
▼ is or is not_________ exceptionally small.

Thanks!

bobbym · 2016-02-01 14:33:18

Hi;

8a) I am getting .5460

8b) I am getting .9936

8c) Yes, because the probability of getting a sample mean of 0.8543 or greater when 464 candies are selected is not exceptionally small.

sassytonigirl · 2016-02-01 15:41:20

Can you tell me how to set the problems for #4 and #8 up??

Thanks!

bobbym · 2016-02-01 17:18:18

I can not explain it any better than this:

https://www.wyzant.com/resources/answer … his_please

sassytonigirl · 2016-02-02 08:41:27

Thank you for your help....SO THANKFUL MY CAREER IS NURSING AND NOT THIS MATH...I do not understand all this stuff!!!!!

sassytonigirl · 2016-02-02 08:42:34

bob..can you help me set up 4c

Bob · 2016-02-02 20:58:22

hi sassytonigirl

It would be impossible to have a normal distribution table for every mean and standard deviation, so there is just one. It has a mean of zero and a standard deviation of 1. This diagram shows the men's height curve being converted into the standard curve.

You take every z, subtract the mean, get the result and divide by the sd. The result is a new normal curve with a mean of zero and sd = 1.

To exclude the tallest 5% that green shaded area must be 0.95. So if phi is the function that converts a 'z' into a probability (of being less than z)

Because the normal curve is symmetrical, tables usually only give one half of the possible values to save space. The MIF table here:

http://www.mathsisfun.com/data/standard … table.html

starts at z=0 and just gives the probability from there onwards, so you'd use it by looking up 0.45 (because 0.5 + 0.45 = 0.95).

Before you get to the table on that page you will find an interactive graph. The default matches the table, but, on the left are three buttons for changing the way the interaction works. If you click for "Up to Z" then the shaded area can be adjusted to show the probability you want. The closest I can get by adjusting the line is 95.05 when the Z value is 1.65. I can improve the accuracy a little by looking at the table. 0.4495 gives Z = 1.64 and 0.4505 gives Z = 1.65 so I'd go halfway and say 0.45 is Z = 1.645.

So I have

and you can solve for z from that. That gives you the maximum allowable height for men.

The women's problem can be done in the same way but with a new mean and sd. Use the symmetry of the curve, so that you're working on the left hand side of zero to get the lowest height here.

Bob

Last edited by Bob (2016-02-02 21:01:13)

Relentless · 2016-02-02 21:23:12

Hello,

4a. A long fraction that is about 98.4613432450151494090841495823956779076535639205758343450% (Only about 1 in 4 million women are too tall)

4b. A long error function expression that is about 97.9033457904731490232418820959575107681464529025371879797%

For 4c., you have to use a table to find the z-score (multiple of standard deviation) from the percentage/probability instead of vice versa. I cannot find an exact value for the z-score of the bottom 95%, but it is around 1.64485362513. So we multiply this number by the standard deviation, and that is about how far above/below the mean the maximum/minimum lies.
For the top 5% of males, multiply this number by the male s.d. and add it to the male mean. We get a maximum of about 6ft 1.87in.
For the bottom 5% of females, multiply this number by the female s.d. and subtract it from the female mean. We get a minimum of about 4ft 10.29in.

Last edited by Relentless (2016-02-02 21:25:44)

Relentless · 2016-02-02 21:47:16

7a. An erf expression that is about 62.3918845810973043106238234927825020672554920384931% I would expect a bit of error in a typical answer because you would not normally find the exact conversion for 6/19 of a s.d.

Similarly for 8a. because you would not normally convert exactly -59/511 of a s.d. The answer is about 54.595969158773076487246800747118403970727804830356%

Last edited by Relentless (2016-02-02 21:49:08)

Math Is Fun Forum

#1 2016-01-31 12:48:39

****Can U Help Me Set These Up & What You Get For Comparison To Mine

#2 2016-01-31 13:29:57

Re: ****Can U Help Me Set These Up & What You Get For Comparison To Mine

#3 2016-01-31 14:14:39

Re: ****Can U Help Me Set These Up & What You Get For Comparison To Mine

#4 2016-01-31 17:15:32

Re: ****Can U Help Me Set These Up & What You Get For Comparison To Mine

#5 2016-01-31 23:28:29

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#6 2016-02-01 00:57:13

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#7 2016-02-01 04:06:01

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#8 2016-02-01 11:45:01

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#9 2016-02-01 14:33:18

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#10 2016-02-01 15:41:20

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#11 2016-02-01 17:18:18

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