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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?

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