Results 1 to 2 of 2

Math Help - statistics % aged less than 24.5 using mean 30 and standard deviation of 3

  1. #1
    namethattuneinone
    Guest

    statistics % aged less than 24.5 using mean 30 and standard deviation of 3

    Hi all totally new here. I have not done maths for thirty years and find myself stuck with this one.

    The ages of all new mothers have a mean of 30.0 years and a standard deviation of 3.0 years.

    Provide answers correct to one decimal place below (e.g. 4.0 for 4.0%).
    Assuming that the data is normally distributed, the percentage of new mothers aged less than 24.5 years is % in the population.

    Using 95% rule of 2 standard deviations either side of the mean mothers less than 24 would = 2.5%. But what do you do when you are asked to work it out to something that does not fit the stasndar deviation. i.e. 24.5

    Thanks in advance for any help
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member teuthid's Avatar
    Joined
    Feb 2008
    From
    Bowling Green, OH
    Posts
    49

    Z-score

    Calculating probabilities with normal distributions directly is extremely difficult (the formula is P(x\leq X)=\int_{-\infty}^{x} \frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}}dz. What's worse is that every unique situation (weights of dogs, height of grass, lifespan of light bulbs, grades in statistics, etc.) would require a unique set of calculations.
    So in order to save everybody some time, the probability calculations for a normal distribution with a mean of 0 and a standard deviation of 1 have been calculated and are readily available in tables (there's probably one in the back of your textbook). All we have to do is to convert the information from our normal distribution to the standard normal distribution. Here's the formula:

     z=\frac{X-\mu}{\sigma}

    The X represents the particular data point from your population. The \mu represents the mean of the population, and the \sigma represents the standard deviation. The z is the converted data value for the standard normal distribution.

    Since this question starts with a data point (mothers younger than 24.5) and asks for a percentage, we will begin by calculating the z-score of our data point:

     z=\frac{X-\mu}{\sigma}=\frac{24.5-30.0}{3.0}=-1.8

    Now we need to go find a table of z-scores and look up the percentage that is associated with a z-score of -1.8. The table that I used listed the value as .0359 or 3.59%. To one significant digit, the percentage would be 4.0%

    This method will work for any data point you need to test, as long as the data in question is normally distributed.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 9
    Last Post: May 20th 2011, 09:29 PM
  2. Replies: 5
    Last Post: May 18th 2009, 04:14 AM
  3. standard deviation and mean deviation
    Posted in the Statistics Forum
    Replies: 1
    Last Post: March 16th 2008, 05:09 AM
  4. Statistics...standard deviation
    Posted in the Statistics Forum
    Replies: 2
    Last Post: December 1st 2007, 08:00 AM
  5. statistics-deviation
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: January 10th 2007, 02:52 PM

/mathhelpforum @mathhelpforum