In: Statistics and Probability
Solution:
Given that,
Y~N(480, 99)
μ =480,σ=√99 =9.9499
A) let ' a ' be the the 75th percentile.
P( Y< a)= 0.75
p{[(Y- μ)/σ]<[(a - 480)/9.9499]}=0.75
P( Z< z)= 0.75 .... where z=(a - 480)/9.9499
By using the standard normal table , we get
P( Z< 0.67)= 0.75
z= 0.67
Now using the z-score formula, we can find the value of ' a ' as
follows,
z=(a- μ)/σ
z=(a - 480)/9.9499
(0.67)×9.9499=( a - 480)
a= 480+[(0.67)×9.9499]
=480 + 6.6664
a=486.6664
P( Y< 486.6664)= 0.75
75th percentile is 486.6664
B) let ' a ' be the the 25th percentile.
P( Y< a)= 0.25
p{[(Y- μ)/σ]<[(a - 480)/9.9499]}=0.25
P( Z< z)= 0.25 .... where z=(a - 480)/9.9499
By using the standard normal table , we get
P( Z< -0.67)= 0.25
z= -0.67
Now using the z-score formula, we can find the value of ' a ' as
follows,
z=(a- μ)/σ
z=(a - 480)/9.9499
(-0.67)×9.9499=( a - 480)
a= 480+[(-0.67)×9.9499]
=480−6.6664
a=833.3336
P( Y< 833.3336)= 0.25
25th percentile is 833.3336