Question

Sketch the area under the standard normal curve over the indicated interval and find the specified area. (Round your answer to four decimal places.)

A) The area to the left of z = 0.79 is

B) The area to the right of z = −2.10 is

C) The area between z = 0 and z = −1.99  is

D) The area between z = −1.31 and z = 1.92 is

Answer 1

Φ(a) is the cumulative probability distribution which is equal to P(Z<a)

1) area to the left of 0.79

which is the P(Z < 0.79)=Φ(0.79)

The area to the left of z = 0.79 is= 0.7852

2)  area to the right of z = −2.10 is

which is the

P(Z > -a)

The probability of P(Z > –a) is P(a), which is Φ(a)= Φ(2.10)=0.9821

remember

P(Z<a)= Φ(a) and also  Φ(a) + Φ(–a) = 1

∴ Φ(–a) = 1 – Φ(a)

3)

P(a < Z < b) = P(Z < b) – P(Z < a),

P(Z < b) – P(Z < –a) = Φ(b) – Φ(–a)

= Φ(b) – {1 – Φ(a)}

SO,

P(-1.99 < Z < 0) = Φ(0) – {1 – Φ(-1.99)} =0.5-{1 – 0.0233}

The area between z = 0 and z = −1.99  is = 0.4767

4)

similarly as above ,

P(a < Z < b) = P(Z < b) – P(Z < a),

P(Z < b) – P(Z < –a) = Φ(b) – Φ(–a)

= Φ(b) – {1 – Φ(a)}

P(-1.31 < Z < 1.92) = Φ(1.92) – {1 – Φ(1.31)} =0.97257 - {1 – 0.9049}

Thus area between z = −1.31 and z = 1.92 is= 0.8775