Let Z be a standard normal random variable and calculate the following probabilities, drawing pictures wherever appropriate. (Round your answers to four decimal places.)(a) P(0 <= Z <= 2.24)(b) P(0 <= Z <= 2)(c) P(%u22122.60 <= Z <= 0)(d) P(%u22122.60 <= Z <= 2.60)(e) P(Z <= 1.64)(f) P(%u22121.75 <= Z)(g) P(%u22121.60 <= Z <= 2.00)(h) P(1.64 <= Z <= 2.50)(i) P(1.60 <= Z)(j) P(|Z| <= 2.50)

(a) P(0 <= Z <= 2.24)  = 0.4927

(b) P(0 <= Z <= 2)  = 0.4773

(c) P(-2.60 <= Z <= 0) = 0.4953

(d) P(-2.60 <= Z <= 2.60) = 0.9906

(e) P(Z <= 1.64)  = 0.9495

(f) P(-1.75 <= Z) = 0.0047

(g) P(-1.60 <= Z <= 2.00) =  0.9425

(i) P(1.60 <= Z)=0.0548

(j) P(|Z| <= 2.50) = 0.9876

Step-by-step explanation:

(a) P(0 <= Z <= 2.24) = P(Z <= 2.24)- P(Z <= 0)

using the STANDARD NORMAL DISTRIBUTION TABLE

P(0 <= Z <= 2.24) = 0.9927 -  0.5  = 0.4927

(b) P(0 <= Z <= 2) = P(Z <= 2)- P(Z <= 0)

= 0.9773 -  0.5  = 0.4773

(c) P(-2.60 <= Z <= 0) = P(Z <= 0)- P(-2.60)

=   0.5 - 0.0047 = 0.4953

(d) P(-2.60 <= Z <= 2.60) = P(Z <= 2.6)- P(-2.60)

=   0.9953 - 0.0047 = 0.9906

(e) P(Z <= 1.64)  = 0.9495

(f) P(-1.75 <= Z) =1 - P(Z < 2.6) = 1 -  0.9953  = 0.0047

(g) P(-1.60 <= Z <= 2.00) = P(Z <= 2.0)- P(-1.60)

= 0.9773- 0.0548  = 0.9425

(i) P(1.60 <= Z)=1 - P(Z < 1.6) = 1 -  0.9452  = 0.0548

(j) P(|Z| <= 2.50) = P(-2.5 < Z <= 2.50)= P(Z <= 2.5)- P(-2.5)

=0.9938 - 0.0062 = 0.9876

f(x) = x^3 -2x^2 -3x

x( x^2 -2x-3)

x( x^2 +x -3x -3)

x( x(x+1) -3(x+1))

x( x-3)( x+1)

so zeroes are 0, 3,-1

its (2) option