- #1
Aftermarth
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ok. mean ([tex]\mu\[/tex]) and standard deviation ([tex]\sigma\[/tex]) are unknown.
20% of people scored less than 45
and the top 15% scored greater than 87
thus:
P(x [tex]\leq\[/tex] 45) = .2
P(x > 87) = 0.15, which needs to be converted to P(x [tex]\leq\[/tex] 87 ) = 0.85
now using z scores ( z - [tex]\mu\[/tex]) / [tex]\sigma\[/tex]
for part one:
(45 - [tex]\mu\[/tex]) / [tex]\sigma\[/tex] = inverse normal (0.2)
= -0.8416...
rearranging to make 45 the subject:
-0.8416[tex]\sigma\[/tex] + [tex]\mu\[/tex] = 45
and for part 2:
(87 - [tex]\mu\[/tex]) / [tex]\sigma\[/tex] = inverse normal (0.85)
= 1.03643...
rearranging to make 87 the subject:
1.03643[tex]\sigma\[/tex] + [tex]\mu\[/tex] = 87
this leaves to simulataneous equations:
-0.8416[tex]\sigma\[/tex] + [tex]\mu\[/tex] = 45
1.03643[tex]\sigma\[/tex] + [tex]\mu\[/tex] = 87
which can be solved to give:
[tex]\mu\[/tex] = 63.8
[tex]\sigma\[/tex] = 22.4
am i correct?
20% of people scored less than 45
and the top 15% scored greater than 87
thus:
P(x [tex]\leq\[/tex] 45) = .2
P(x > 87) = 0.15, which needs to be converted to P(x [tex]\leq\[/tex] 87 ) = 0.85
now using z scores ( z - [tex]\mu\[/tex]) / [tex]\sigma\[/tex]
for part one:
(45 - [tex]\mu\[/tex]) / [tex]\sigma\[/tex] = inverse normal (0.2)
= -0.8416...
rearranging to make 45 the subject:
-0.8416[tex]\sigma\[/tex] + [tex]\mu\[/tex] = 45
and for part 2:
(87 - [tex]\mu\[/tex]) / [tex]\sigma\[/tex] = inverse normal (0.85)
= 1.03643...
rearranging to make 87 the subject:
1.03643[tex]\sigma\[/tex] + [tex]\mu\[/tex] = 87
this leaves to simulataneous equations:
-0.8416[tex]\sigma\[/tex] + [tex]\mu\[/tex] = 45
1.03643[tex]\sigma\[/tex] + [tex]\mu\[/tex] = 87
which can be solved to give:
[tex]\mu\[/tex] = 63.8
[tex]\sigma\[/tex] = 22.4
am i correct?
Last edited: