Ask Question
12 October, 13:34

Determine the area under the standard normal curve that lies between (a) z=-1.36 and z=1.36 (b) z=-1.98 and z=0, and (c) = 1.14 and z=-0.33. The area that lies between z=-1.36 and z=1.36 is

+5
Answers (1)
  1. 12 October, 15:14
    0
    (a) The area that lies between z=-1.36 and z=1.36 is 0.82616

    (b) The area that lies between z=-1.98 and z=0 is 0.47615

    (c) The area that lies between z=1.14 and z=-0.33 is 0.50215

    Step-by-step explanation:

    Determine the area under the standard normal curve that lies between:

    (a) z=-1.36 and z=1.36

    This is:

    P (-1.36<=z<=1.36) = P (z<=1.36) - P (z<=-1.36)

    P (z<=1.36) = 0.91308

    P (z=1.36) = 1-P (z<=1.36) = 1-0.91308→P (z<=-1.36) = 0.08692

    P (-1.36<=z<=1.36) = P (z<=1.36) - P (z<=-1.36) = 0.91308-0.08692→

    P (-1.36<=z<=1.36) = 0.82616

    (b) z=-1.98 and z=0

    This is:

    P (-1.98<=z<=0) = P (z<=0) - P (z<=-1.98)

    P (z<=0) = 0.5

    P (z=1.98) = 1-P (z<=1.98)

    P (z<=1.98) = 0.97615

    P (z<=-1.98) = 1-P (z<=1.98) = 1-0.97615→P (z<=-1.98) = 0.02385

    P (-1.98<=z<=0) = P (z<=0) - P (z<=-1.98) = 0.5-0.02385→

    P (-1.98<=z<=0) = 0.47615

    (c) z=1.14 and z=-0.33

    This is:

    P (-0.33<=z<=1.14) = P (z<=1.14) - P (z<=-0.33)

    P (z<=1.14) = 0.87285

    P (z=0.33) = 1-P (z<=0.33)

    P (z<=0.33) = 0.62930

    P (z<=-0.33) = 1-P (z<=0.33) = 1-0.62930→P (z<=-0.33) = 0.3707

    P (-0.33<=z<=1.14) = P (z<=1.14) - P (z<=-0.33) = 0.87285-0.3707→

    P (-0.33<=z<=1.14) = 0.50215
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “Determine the area under the standard normal curve that lies between (a) z=-1.36 and z=1.36 (b) z=-1.98 and z=0, and (c) = 1.14 and ...” in 📘 Mathematics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers