Question

 

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

Answer 1

(a)Given P[Z < c] = 0.8980

P[Z < c] =P[-Z < c]=0.8980

  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[-Z 0]=0.992

  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[-Z < c]=0992

  P[0Z < c]=.992-0.5=0.492

c=2.333

  (g) P[|Z| < c] = 0.4246

P[-cZ 0]+P[0Z < 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[-cZ 0]+P[0Z < c]= 0.984

=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[-cZ 0]+P[0Z < c]= .204

=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