Find the area under the standard normal curve which lies

Question

A. to the left of 0.47

  b) to the right of z = -0.06

  ( c) to the right of z = 2.53

d) to the left of z = -1.31

  (e) between z = 0.72 and z = 1.59

  (f) between z = -0.11 and z = 2.48

Answer 1

(a) For z = 0.47, I have a Casio calculator program that will give highly accurate A equal to 0.1808224912. Take Area to the left of z = 0.47 then as 0.5 + 0.1808224912 or 0.6808224912 equivalent to 0.6808.

(b) For z = -0.06, my program gives A as |-0.02392218265|. Take Area to the right of z = -0.06 then as |-0.02392218265| + 0.5 or |0.52392218265| equivalent to 0.5239.

(c) For z = 2.53, my program gives A as 0.4942968737. Take Area to the right of z = 2.53 then as

0.5 − 0.4942968737 or 0.005703126335 equivalent to 0.0057.

(d) For z = -1.31, my program gives A as |-0.4049020822|. Take Area to the left of z = -1.31 then as

|-0.5| − |-0.4049020822| or 0.0950979178 equivalent to 0.0951.

(e) For z = 0.72, my program gives A as 0.2642375022. For z = 1.59, my program gives A as 0.4440825975. Take Area between z = 0.72 and z = 1.59 then as 0.4440825975 − 0.2642375022 or 0.1798450953 equivalent to 0.1798.

(f) For z = -0.11, my program gives A as |-0.04379531254|. For z = 2.48, my program gives A as 0.4934308809. Take Area between z = -0.11 and z = 2.48 then as |-0.04379531254| + 0.4934308809

or 0.5372261934 equivalent to 0.5372.