The amount of time it takes to recover physiologically from a certain kind of su

Question

The amount of time it takes to recover physiologically from a certain kind of sudden noise is found to be normally distributed with a mean of 80 seconds and a standard deviation of 10 seconds. Using the 50%-34%-14% figures , approximately what percentage of scores (on time to recover) will be (a)above .100, (b) below 100, (c) above90, (d) below 90, (e) above 80, (f) below 80, (g) above 70, and (h) below 70 , (i) above 60, and (j) below 60?


The amount of time it takes to recover physiologically from a certain kind of sudden noise is found to be normally distributed with a mean of 80 seconds and a standard deviation of 10 seconds. Using the 50%-34%-14% figures , approximately what percentage of scores (on time to recover) will be (a)above .100, (b) below 100, (c) above90, (d) below 90, (e) above 80, (f) below 80, (g) above 70, and (h) below 70 , (i) above 60, and (j) below 60?


The amount of time it takes to recover physiologically from a certain kind of sudden noise is found to be normally distributed with a mean of 80 seconds and a standard deviation of 10 seconds. Using the 50%-34%-14% figures , approximately what percentage of scores (on time to recover) will be (a)above .100, (b) below 100, (c) above90, (d) below 90, (e) above 80, (f) below 80, (g) above 70, and (h) below 70 , (i) above 60, and (j) below 60?

Explanation / Answer

Mean = 80

Stdev = 10

a. P(X>100) = P(Z> 100-80 / 10) = P(Z>2) = 1-.9772 = .0228

b. P(X<100) = P(Z<100-80/10 ) = P(Z<2) = .9772

c. P(X>90) = P(Z> 90-80/10) = P(Z>1) = 1-.8413 = .1587

d. P(X<90) = P(Z<1) = .8413

e. P(above 80) = P(Z>80-80/10) = P(Z>0) = .50

f. P(below 80) = P(Z<80) = 1-P(Z>80) = .5

g. P(above 70) = P(Z>70-80/10) = P(Z>-1) = 1-.1587 = .8413

h. P(below 70) = 1-P(above 70) = .1587

i. P( above 60) = P(Z>60-80/10) = P(Z>-2) = .9772

j. P(below 60) = P(X<60) = P(Z< 60-80/10) = P(Z<-2) = .0228

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