Aparticular iq test is standardized to a normal model, with a mean of 100 and a standard deviation of 15. a) choose the model for these iq scores that correctly shows what the 68-95-99.7 rule predicts about the scores.

1) 0.0099

2) 0.3707

3) 0.4920

Step-by-step explanation:

Since Mean M = 100

Standard deviation S = 15

1) When the IQ score is x= 68

The percentage of that score up to 68 is normally distributed.

P(z<x) = P(z<68)

P(z<68) = P[ (x-M)/S < (68-100)/15]

P(z< -2.13) = 0.0099

This means that about 1 of every 100 will have IQ score of 68 and below.

2) when the IQ score x = 95

P(z<x) = P(z<95)

P(z<95) = P[ (x-M)/S < (95-100)/15]

P(z< -0.33) = 0.3707

This means that about 37 of every 100 have an IQ score of 95 and below.

3) when the IQ score is x = 99.7

P(z<x) = P(z<99.7)

P(z<99.7) = P[ (x-M)/S < (99.7-100)/15]

P(z< -0.02) = 0.4920

This means that 49 of every 100 have an IQ score of 99.7 and below.

6mm^3 or .6cm^3

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  36     6

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6     6   6

factor trees are nice.