Solutions PS III
1. Let Ebe expenditures
E= 80
E= 10 )2
E= 100
EsN(80;100)
Let Ebe the target budget level. We want Esuch that:
P(EE) = 0:90
)P(EE
EEE
E) = P(zE80
10 ) = 0:90:Let z=E80
10
From the ztable, the value of zthat corresponds to this probability is
z= 1:285
)z=E80
10 = 1:285
)E= 1:285(10) + 80 = 92:85
With this budget, Bob will spend less that this amount 90% of the time.
2. Let the credit score SsN(600;1002)
(i) Let Sbe the value of the credit score that 75% of the customers score
above. We want Ssuch that
P(SS) = 0:75
)P(SS
SS600
100 ) = 0:75
)P(zz) = 0:75;where z=S600
100
)1P(zz) = 1 0:75 = 0:25
The corresponding value of zfrom the ztable is 0:68
)S600
100 =0:68
)S= 600 68 = 532
75% of the applicants will have a credit score exceeding 532.
(ii) For this problem, we consider 80% of the area symmetric about the mean.
P(SLSSU) = 0:80
)P(SSU)P(SSL) = 0:80
)P(zSU600
100 )P(zSL600
100 ) = 0:80
Now since we are looking at an 80% area symmetric about the mean, and
since the normal is a symmetric distribution,
P(zSU600
100 ) = P(zSL600
100 ) = 0:10
Working with the lower limit, let SL600
100 =z:From the ztables, the cor-
responding value of zsuch that P(zz) = 0:10 is 1:28
)z=SL600
100 =1:28
)SL= 600 128 = 472
For the upper limit:
P(zSU600
100 ) = 0:10 )P(zSU600
100 ) = 0:90
Let z=SU600
100 . Then the corresponding value of zfrom the table is 1:28
)SL= 600 + 128 = 728
3. Let Tbe the score
E(T) = 25; (T) = 6; 2
Tn= 36
1
