In: Statistics and Probability
Let Z ∼ N (0, 1) be a standard normal random variable. Find c by using only one side of the z-table for all questions:
(a) P[Z < c] = 0.8980 (b) P[Z > −c] = 0.8980 (c) P[Z > c] = 0.102 (d) P[Z > c] = 0.992
(e) P[Z < −c] = 0.992 (f) P[Z < c] = 0.992 (g) P[|Z| < c] = 0.4246 (h) P[|Z| > c] = 0.1528
(i) P[|Z| < c] = 0.984 (j) P[|Z| > c] = 0.984 (k) P[|Z| < c] = 0.204 (l) P[|Z| > c] = 0.204
(a)Given P[Z < c] = 0.8980
P[Z < c] =P[-
P[0Z < c]=.898-0.5=0.3980.refer the table attached
c=1.27
(b) P[Z > −c] = 0.8980
P[Z > −c] = P[-cZ 0]+P[0Z < ]=0.8980
P[-cZ 0]=.898-0.5=0.3980.
-c=-1.27,c=1.27
(c) P[Z > c] = 0.102
P[Z > c]=o.5-P[0Z < c]=0.102
P[0Z < c]=0.5-.102=0.3980
c=1.27
(d) P[Z > c] = 0.992
P[Z > c]=P[0Z < c]+P[0Z < ]=0.992
P[0Z < c]=0.992-0.5=0.4920
z | 2.33 | 2.34 | .01 |
area | .4901 | .4904 | .0003 |
oneunit distance=.0003/.01=.003, for .4902,2.33+.003=2.333
c=-2.333
(e) P[Z < −c] = 0.992
P[Z <- c] =P[-
P[0Z <- c]=.0.992-0.5=0.4920
-c=2.3333, c=-2.333
(f) P[Z < c] = 0.992
P[Z < c] =P[-
P[0Z < c]=.992-0.5=0.492
c=2.333
(g) P[|Z| < c] = 0.4246
P[-c
=2*P[0Z < c]=0.4246
P[0Z < c]=0.2123
c=0.56
(h) P[|Z| > c] = 0.1528
P[|Z| > c]= P|Z <- c]+P|Z > c]=0.1528
=2*P|Z > c]=0.1528,P|Z > c]=.0764
P|Z > c]=o.5-P[0Z < c]=.0764
P[0Z < c]=0.5-.0764=0.4236
c=1.43
(i) P[|Z| < c] = 0.984
P[-c
=2*P[0Z < c]= 0.984 (since normal curve is symmetric)
P[0Z < c]=0.4920
c=2.333
(j) P[|Z| > c] = 0.984
P[|Z| > c]=P[Z <- c]+=P[Z >c]=0.984
2*P[Z >c]=0.984 ,
P[Z >c]=0.4920,
P[0Z < c]=0.5-0.492=0.108
c=0.38
(k) P[|Z| < c] = 0.204
P[-c
=2*P[0Z < c]= 0.204
P[0Z < c]=0.102
c=0.26
(j) P[|Z| > c] = 0.984 ,P[|Z| > c]=1-P[|Z| < c]=.984
P[|Z| < c]=.016,P[-c 2*P[0Z
< c]=0.016,P[0Z
< c]=.008 c=.02 (l) P[|Z| > c] = 0.204,P[|Z| > c]=1-P[|Z|
< c]=.204 P[|Z| < c]=.796,P[-c 2*P[0Z
< c]=0.796,P[0Z
< c]=.396 c=1.26