the heights of female students at a particular college are normally distributed with the mean 169 cm and standard deviation 9 cm.
1) Given that 80% of these female students have a height less than h cm ,find the value of h.
2) Given that 60% of these female students have a height greater than s cm,find the value of s.
2 answers
0.7
1) H~N(169,9^2)
(H<h)=0.80
(H< (h-169)/9)=0.80
(H<a)=0.80
(fi)a =0.80 (look for 0.80 in the normal distribution table)
the closest # to 0.80 is 0.7995 at 0.84
0.0005 is left to make 0.80 so look for that in the add part of the table 5 is at 2
so the a=0.842
we suggested a as (h-169)/9 =0.842
solve, h=176.6
2) (H>s)=0.6
Standardize the s , (H>(s-169)/9)=0.6
(H>b)=0.6
s is negative because ; if its greater than the random variable which is (s) and the probability is greater than 0.5 then the random variable is negative
rule states that ; (fi)b=0.6
from the table (0.6) is 0.5987 at 0.25
0.0013 left for it to equal 0.6
from the table take the minimum value which is 12 at 3
so b is 0.253
its ngative so b is -0.253
(s-169)/9=-0.253
s=166.7
(H<h)=0.80
(H< (h-169)/9)=0.80
(H<a)=0.80
(fi)a =0.80 (look for 0.80 in the normal distribution table)
the closest # to 0.80 is 0.7995 at 0.84
0.0005 is left to make 0.80 so look for that in the add part of the table 5 is at 2
so the a=0.842
we suggested a as (h-169)/9 =0.842
solve, h=176.6
2) (H>s)=0.6
Standardize the s , (H>(s-169)/9)=0.6
(H>b)=0.6
s is negative because ; if its greater than the random variable which is (s) and the probability is greater than 0.5 then the random variable is negative
rule states that ; (fi)b=0.6
from the table (0.6) is 0.5987 at 0.25
0.0013 left for it to equal 0.6
from the table take the minimum value which is 12 at 3
so b is 0.253
its ngative so b is -0.253
(s-169)/9=-0.253
s=166.7
1 answer