Ask your own question, for FREE!
Mathematics 8 Online
OpenStudy (anonymous):

Let X be a normal random variable with its mean equal to 40 and standard deviation equal to 5.find the following probablilties for this normal distribution a)P(x>55) B)P(x<49)

OpenStudy (unklerhaukus):

find the z scores

OpenStudy (anonymous):

i did N i got 3

OpenStudy (anonymous):

in my data booklet there is no z score at the value of three

OpenStudy (unklerhaukus):

3?

OpenStudy (anonymous):

yes

OpenStudy (unklerhaukus):

each z score is one standard deviation from the mean

OpenStudy (anonymous):

i don't get you...how?

OpenStudy (kropot72):

\[z=\frac{X-\mu}{\sigma}=\frac{55-40}{5}=3\]

OpenStudy (anonymous):

but then @kropot72 in a data booklet there is not a score with the value of 3

OpenStudy (kropot72):

Use the empirical rule: About 99.73% of the area of the normal distribution is included within a distance of plus and minus 3 standard deviations from the mean.

OpenStudy (anonymous):

i've never heard of that formula before..pls show me what you mean.

OpenStudy (zarkon):

@jazrio you must have a weird table if you don't have a value of 3 in your table

OpenStudy (zarkon):

a lot of the tables go up to 3.49

OpenStudy (anonymous):

but @Zarkon my teacher gave me the table and she said that only this values will be given the rest you'll have to find out....ex:the negative z scores

OpenStudy (anonymous):

my one stops at 2.9 only

OpenStudy (zarkon):

you only have half of the normal table?

OpenStudy (zarkon):

weird

OpenStudy (zarkon):

kropot72 method will give you the answer then

OpenStudy (anonymous):

but i dunno how to do it

OpenStudy (anonymous):

can you help me pls

OpenStudy (zarkon):

it would be weird to show you a way that your instructor had not taught.

OpenStudy (anonymous):

yeah but @Zarkon you see my instructor gave me this question long time ago but i don't recall which method she used

OpenStudy (kropot72):

100 - 99.73 = 0.27% of the area lies outside the plus and minus 3 standard deviations from the mean. We are interested in only the upper tail of the normal distribution curve. Therefore the value 0.27% must be halved.

OpenStudy (anonymous):

@kropot72 sorry...i can't visualise

OpenStudy (kropot72):

BTW I have a z table that goes up to z = 4. That table gives 0.1% for z = 3.

OpenStudy (kropot72):

|dw:1340601352114:dw|

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!