Question

Determine the probability of having an error between

a) -1.1 σ and +1.5 σ

b) -2.1 σ and 0 σ

for a normal variable.

[20%) II. Probability for Normal distribution Determine the probability of having an error between a)-1.1 σ and +1.5 σ b)-21 σ and 0 σ for a normal variable. Use Table D.1 from the lecture notes for the standard normal distribution calculations, or use Excel functions.

Standard Normal (Z) Table for positive t Table D.1 Percentage Points for the Standard Normal Distribution Function (continued) 0.0 0.50000 0.50399 0.50798 0.51197 0.51595 0.51994 0.52392 0.52790 0.53188 0.5358 0.1 0.53983 0.54380 0.54776 0.55172 055567 0.55962 0.56356 0.56749 0.57142 0.5753: 0.20.57926 0.58317 0.58706 0.59095 059483 0.59871 0.60257 0.60642 0.61026 0.6140 0.3 0.61791 0.62172 0.62552 0.62930 0.63307 0.63683 0.64058 0.64431 0.64803 0.6517 0.40.65542 0.65910 0.66276 0.66640 0.67003 0.67364 0.67724 0.68082 0.68439 0.6879 5 0.69146 0.69497 0.69847 0.70194 0.70540 0.70884 0.71226 0.71566 0.71904 0.7224 0.6 0.72575 0.72907 0.73237 0.73565 0.73891 0.74215 0.74537 0.74857 0.75175 0.7549 0.70.75804 0.76115 0.76424 0.76730 0.77035 0.77337 0.77637 0.77935 0.78230 0.7852 0.80.78814 0.79103 0.79389 0.79673 0.79955 0.80234 0.80511 0.80785 0.81057 0.8132 0.9 0.81594 0.81859 0.82121 0.82381 0.82639 0.82894 0.83147 0.83398 0.83646 0.8389 1.0 0.84134 0.84375 0.84614 0.84849 0.85083 0.85314 0.85543 0.85769 0.85993 0.8621 1 0.86433 0.86650 0.86864 0.87076 0.87286 0.87493 0.87698 0.87900 0.88100 0.8829 1.2 0.88493 0.88686 0.88877 0.89065 0.89251 0.89435 0.89617 0.89796 0.89973 0.9014 13 0.90320 0.90490 0.90658 0.90824 0.90988 0.91149 0.91309 0.91466 091621 0.9177 4 0.91924 0.92073 0.92220 0.92364 0.92507 0.92647 0.92785 0.92922 0.93056 0.9318 15 0.93319 0.93448 0.93574 0.93699 0.93822 0.93943 0.94062 0.94179 094295 0.9440 1.6 0.94520 0.94630 0.94738 0.94845 094950 0.95053 095154 095254 0.95352 0.9544 2 What is area under curve from -o to 1.68?

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Answer #1

Answer:

Determine the probability of having an error between

  1. -1.1 σ and +1.5 σ

Using standard normal distribution,

we have to find P( -1.1<z<1.5)

= P( z <1.5)-P( z < -1.1)

Excel function: =NORM.S.DIST(1.5,TRUE) -NORM.S.DIST(-1.1,TRUE)

Answer: =0.7975

  1. -2.1 σ and 0 σ

we have to find P( -2.1<z<0)

= P( z <0)-P( z < -2.1)

= 0.4821

Excel function: =NORM.S.DIST(0,TRUE) -NORM.S.DIST(-2.1,TRUE)

c).

area under the curve –infinity to 1.68

= P( z < 1.68)

= 0.953521

Excel function: =NORM.S.DIST(1.68,TRUE)

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