a group of 234 students has a mean age of 14.2 with a standard deviation of .2 years. the ages are normally distributed. how many students are younger than 14.3 years? express answer to the nearest student.
@amistre64
lets move this whole setup so that the mean zeros out by subtracting the mean from every point x x - mean to determine the variation in the spread; lets see how many standard deviations fit within the spread: (x-mean)/sd this gives us a standardized score for all the values within a normal distribution
ughm ok
so, to determine this standard score (z score) take the value they give you and subtract the mean
14.3-14.2?
yes
.1
now, lets see how many standard deviations fit into the range divide that by the value of the standard deviation they give you
.2/.1
=2
.1 divided by .2 = 1/2
oops .5
there are tables created, or calculators constructed, that relate this zscore to the area under the normal distribution curve
the tables can be iffy since each author perfers a different style
do you have a ti83 by chance?
............. r u asking me to make a table?
no i have a TI-30XA
make a table? lol, no back in the ancient days, they built tables to record things with. This prevented a lot of tediuos chore of reevaluating the same value over and over and over again
not sure if a financial calculator is sufficient for a normalCDF function
ohhhhhh okay so what are you asking me to do next
without a calculator, or a program to assist you, you will need to find a suitable z table to look in
one should be provided to you with your course material
do you by chance have a list of options to choose from?
its not all i have is this worksheet
i tend to like to draw the humpback and shade in the setup we are looking for: |dw:1369146051582:dw| so we want at best, a left tailed z table; otherwise we would have to do some extra adjustments to fit
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