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Mathematics 15 Online
OpenStudy (shamuta):

STATS Question~! Let X be N(μ= 25, (σ^2)= 16). Determine the following probabilities: (a.) P(22< X <26) (b.) P(19< X<24) (c.) P(X>23) (d.) P(|X-25| <5)

OpenStudy (cwtan):

standard deviation =4 (a) P(22<x<26)= P(-0.75<Z<0.75)= 1-2P(Z>0.75) get P(Z>0.75) and u found part A

OpenStudy (shamuta):

where did the -.75 and the 0.75 come from?

OpenStudy (cwtan):

Z= \(\Huge \frac {x-μ}{σ}\)

OpenStudy (cwtan):

\(\Huge \frac {22-25}{4}\) and \(\Huge \frac {26-25}{4}\) oops it's 0.25

OpenStudy (shamuta):

so it would be ((22-25)/4) which gives -0.75, but you show a 0.75, so you show both the value, and it's opposite? unless you mean for the second value to be (26-25)/4= 0.25?

OpenStudy (cwtan):

sorry sorry.....

OpenStudy (shamuta):

oh ok lol XD Thanks a lot~! :D

OpenStudy (shamuta):

oh wait, P(−0.75 ≤ Z ≤ 0.25) = P(Z ≤ 0.25) − P(Z ≤ −0.75) = 0.5987-0.2266 = 0.3721 ^is that the correct answer?

OpenStudy (shamuta):

Let X be N(μ= 25, (σ^2)= 16). Determine the following probabilities: P(|X-25| <5) How would you deal with the absolute value in this part of the question?

OpenStudy (zarkon):

P(|X-25| <5)=P(-5<X-25<5)

OpenStudy (shamuta):

^ah ok, Thanks! :D

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