Results 1 to 3 of 3

Math Help - Statistics questions

  1. #1
    Banned
    Joined
    Apr 2008
    Posts
    1

    Cool Statistics questions

    1. Suppose that in the 2004, 10 percent of the voters are still undecided. Find the probability that among 5 voters questioned, exactly 2 of them are undecided.
    pg190 P(x)=nCx (π)x(1-x) n-x π =.10 x=2 n=5
    P(2)= 5C2(.10) 2 (1-2) 5-2 = ( ) (.01)(-1)
    ^ ^
    ??What is this value and how is it determined

    2. On each SAT math section there are 50 questions, each question has four possible answers, one of which is correct. For students who guess at all answers, find the standard deviation for the number of correct answers.

    pg 191 o2=nπ (1-π) n=50 π=1/4 =50(.25)(1-50)=12.5(-49)=-612.5=612.5
    o2=612.5

    3. Doug leads bird-watching trips every morning in March. The number of cardinals seen has a Poisson distribution with a mean of 2.0. Find the probability that on a randomly selected trip, the number of cardinals seen is 2.
    pg204 P(x)=uXe-u u=2.0 X=2
    X! P(2) = (2.0) 2 (e-2.0) = (4.)(.135) = .54 = .27
    2! 2! 2!

    4. The ACME Scientific Instrument Company manufactures thermometers that are supposed to give readings of 0 degrees Celsius at the freezing point of water. Tests on a large sample of these thermometers reveal that at the freezing point of water, some give readings below 0 degrees (denoted by negative numbers) and some give readings above 0 degrees (denoted by positive numbers). Assume that the mean reading is 0 degrees and the standard deviation of the readings is 1.00 degrees. Also assume that the frequency distribution of errors closely resembles the normal distribution. If 7% of the thermometers are rejected because they have readings that are too high, but all other thermometers are acceptable, find the temperature that separates the rejected thermometers from the others.

    Pg 229 Z=X-u u=0 o=1 pr(X.>x)=0.07 z=X-0 z=.5000-.07=.43
    0 1
    Appendix B1 closest to .4300 is .4292 z value is 1.47
    1.47=X-0 = X= 0- 1.47(1)=-1.47
    1
    5. IQ scores of UIU professors are normally distributed with a mean of 105 and a standard deviation of 21. In a random sample of 90, approximately how many Profs will have IQs between 84 and 133?

    Pg 251 n=90 u=105 o=21 Y=random variables between 84 and 133
    According to excel true 0.24 ?? Please assist me in answering this problem
    6. A study of the amount of time it takes a mechanic to rebuild the transmission for a 1998 Acura Integra shows that the mean is 8.4 hours and the standard deviation is 1.8 hours. If 40 mechanics are randomly selected, find the probability that their mean rebuild time exceeds 8.1 hours. u=8.4 o=1.8/√40
    Pg 282 Z= -u Z= 8.1 -8.4 = -0.3 = -1.05 P=0.1469 found by .5000-0.3531
    o√n 1.8/√40 0.285 ?? SHOULD THIS BE^ ADDED OR SUBTRACTED


    7. Find the probability that in 200 tosses of a pair fair dice, we will obtain at least 40 sevens. Assume that it is a normal distribution.
    Pg 191 o2= n π(1- π) π=40 n=200 02 =200(40)(1-40)=8000(-39)=-312000
    Standard deviation is 558.57



    8. Use the given degree of confidence and sample data to find the margin of error in estimating the population mean μ for the following scenario. College students' annual earnings: 99% confidence; n = 67, = $6068, s = $1000.
    Pg 304
    ± t s df = 67-1=66 t=2.652 = $6068 ± 2.652 $1000 =324.01 ± $6068
    √n √67
    ?? what are The end point of confidence intervals and how did u determine this
    Last edited by Seansgirl33; April 20th 2008 at 11:32 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Jan 2008
    From
    Montreal
    Posts
    311
    Awards
    1
    1)  \binom{5}{2} (0.1)^{2}(0.9)^{3} = 0.0729

    2) I'm not sure of my answer but I think it's:

     \sqrt{(50)^2 (4) \left( \frac{1}{4} \right) \left( \frac{3}{4} \right)} = 43.3012

    3)  \frac{2^2}{2!}e^{-2} = 0.27067

    4) the value associated with 7% is about 1.47.

     1.47 = \frac{x-0}{1} = 1.47 so 1.47 is the cutoff temperature

    5)  \frac{133-105}{21} = 0.0918
     \frac{84-105}{21} = 0.1587
     1-0.0918-0.1587=0.7495

    6)  \frac{8.1-8.4}{1.8} = 0.4325 \Rightarrow 1-0.4325 = 0.5675

    7) I'm not sure but I think it's:

    P(X \geq 40) = 1- P(X<40) = 1-(P(X=0) \cup P(X=1) \cup ... \cup P(X=39)

    P(X=0) = \binom{200}{0} \left( \frac{1}{6} \right)^0 \left( \frac{5}{6} \right)^{200}
    P(X=1) = \binom{200}{1} \left( \frac{1}{6} \right) \left( \frac{5}{6} \right)^{199}
    P(X=39) = \binom{200}{39} \left( \frac{1}{6} \right)^{39} \left( \frac{5}{6} \right)^{161}

    or using excel, by setting up the columns to go from 0-39, then the formula "= Binomdist("the column with the numbers", 200, 0.166667, FALSE)" Again I'm not 100% of this answer

    8. ?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Duplicate post. Thread closed.
    Quote Originally Posted by Seansgirl33 View Post
    1. Suppose that in the 2004, 10 percent of the voters are still undecided. Find the probability that among 5 voters questioned, exactly 2 of them are undecided.
    pg190 P(x)=nCx (π)x(1-x) n-x π =.10 x=2 n=5
    P(2)= 5C2(.10) 2 (1-2) 5-2 = ( ) (.01)(-1)
    ^ ^
    ??What is this value and how is it determined

    2. On each SAT math section there are 50 questions, each question has four possible answers, one of which is correct. For students who guess at all answers, find the standard deviation for the number of correct answers.

    pg 191 o2=nπ (1-π) n=50 π=1/4 =50(.25)(1-50)=12.5(-49)=-612.5=612.5
    o2=612.5

    3. Doug leads bird-watching trips every morning in March. The number of cardinals seen has a Poisson distribution with a mean of 2.0. Find the probability that on a randomly selected trip, the number of cardinals seen is 2.
    pg204 P(x)=uXe-u u=2.0 X=2
    X! P(2) = (2.0) 2 (e-2.0) = (4.)(.135) = .54 = .27
    2! 2! 2!

    4. The ACME Scientific Instrument Company manufactures thermometers that are supposed to give readings of 0 degrees Celsius at the freezing point of water. Tests on a large sample of these thermometers reveal that at the freezing point of water, some give readings below 0 degrees (denoted by negative numbers) and some give readings above 0 degrees (denoted by positive numbers). Assume that the mean reading is 0 degrees and the standard deviation of the readings is 1.00 degrees. Also assume that the frequency distribution of errors closely resembles the normal distribution. If 7% of the thermometers are rejected because they have readings that are too high, but all other thermometers are acceptable, find the temperature that separates the rejected thermometers from the others.

    Pg 229 Z=X-u u=0 o=1 pr(X.>x)=0.07 z=X-0 z=.5000-.07=.43
    0 1
    Appendix B1 closest to .4300 is .4292 z value is 1.47
    1.47=X-0 = X= 0- 1.47(1)=-1.47
    1
    5. IQ scores of UIU professors are normally distributed with a mean of 105 and a standard deviation of 21. In a random sample of 90, approximately how many Profs will have IQs between 84 and 133?

    Pg 251 n=90 u=105 o=21 Y=random variables between 84 and 133
    According to excel true 0.24 ?? Please assist me in answering this problem
    6. A study of the amount of time it takes a mechanic to rebuild the transmission for a 1998 Acura Integra shows that the mean is 8.4 hours and the standard deviation is 1.8 hours. If 40 mechanics are randomly selected, find the probability that their mean rebuild time exceeds 8.1 hours. u=8.4 o=1.8/√40
    Pg 282 Z= -u Z= 8.1 -8.4 = -0.3 = -1.05 P=0.1469 found by .5000-0.3531
    o√n 1.8/√40 0.285 ?? SHOULD THIS BE^ ADDED OR SUBTRACTED


    7. Find the probability that in 200 tosses of a pair fair dice, we will obtain at least 40 sevens. Assume that it is a normal distribution.
    Pg 191 o2= n π(1- π) π=40 n=200 02 =200(40)(1-40)=8000(-39)=-312000
    Standard deviation is 558.57



    8. Use the given degree of confidence and sample data to find the margin of error in estimating the population mean μ for the following scenario. College students' annual earnings: 99% confidence; n = 67, = $6068, s = $1000.
    Pg 304
    ± t s df = 67-1=66 t=2.652 = $6068 ± 2.652 $1000 =324.01 ± $6068
    √n √67
    ?? what are The end point of confidence intervals and how did u determine this
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Two Statistics Questions
    Posted in the Statistics Forum
    Replies: 5
    Last Post: January 31st 2011, 02:51 PM
  2. Please help with statistics questions!
    Posted in the Statistics Forum
    Replies: 0
    Last Post: January 17th 2010, 04:28 PM
  3. statistics questions
    Posted in the Statistics Forum
    Replies: 2
    Last Post: May 20th 2008, 11:57 AM
  4. very last statistics questions
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: January 8th 2008, 03:26 AM
  5. A few statistics questions
    Posted in the Advanced Statistics Forum
    Replies: 13
    Last Post: June 6th 2007, 11:22 AM

Search Tags


/mathhelpforum @mathhelpforum