Asked by sharon
Please help. I have no idea what I am doing
Use the standard normal (z score) table to find: P(-1.00 ≤ z)
&
Find the probability that a data value picked at random from a normal population will have a standard score (z) that lies between the following pairs of z-values.
z = 0 to z = 2.10
Use the standard normal (z score) table to find: P(-1.00 ≤ z)
&
Find the probability that a data value picked at random from a normal population will have a standard score (z) that lies between the following pairs of z-values.
z = 0 to z = 2.10
Answers
Answered by
Reiny
-1.00 ≤ z is the same as z ≥ -1
or
1 - P(z ≤ -1.00) ----> .8413
I don't know how old your tables are or what they look like, but for most of them ...
here is a typical one:
http://www.stat.ufl.edu/~athienit/Tables/Ztable.pdf
find -1.0 in the horizontal under the 00 vertical to get . 1587
so
1 - P(z ≤ -1.00)
= 1 - .1587 = .8413
Of course with internet access, the use of tables is obsolete and we use on-line applets like this one:
http://davidmlane.com/normal.html
since you want the values above a z-score of -1
leave the mean at 0
and the SD at 1
click on "above" and enter -1
read .8413 in the answer window.
or
1 - P(z ≤ -1.00) ----> .8413
I don't know how old your tables are or what they look like, but for most of them ...
here is a typical one:
http://www.stat.ufl.edu/~athienit/Tables/Ztable.pdf
find -1.0 in the horizontal under the 00 vertical to get . 1587
so
1 - P(z ≤ -1.00)
= 1 - .1587 = .8413
Of course with internet access, the use of tables is obsolete and we use on-line applets like this one:
http://davidmlane.com/normal.html
since you want the values above a z-score of -1
leave the mean at 0
and the SD at 1
click on "above" and enter -1
read .8413 in the answer window.
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