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)

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gkosworldreign gkosworldreign · Answer 1 · 2020-03-02T06:14:59+01:00

Answer:

(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