For bone density scores that are normally distributed with a mean of 0 and a standard deviation of 1, find the percentage of scores that area. significantly high (or at least 2 standard deviations above the mean).b. significantly low (or at least 2 standard deviations below the mean).c.

a)

For this case we can use the normal standard table or excel and the complement rule and we got:

b)

For this case we can use the normal standard table or excel and we got:

c)

And we can find this probability with this difference:

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Part a

Let Z the random variable that represent the bone density scores of a population, and for this case we know the distribution for X is given by:

Where and

We are interested on this probability

For this case we can use the normal standard table or excel and the complement rule and we got:

Part b

We are interested on this probability

For this case we can use the normal standard table or excel and we got:

Part c: not significant (or less than 2 standard deviations away from the mean).

For this case we want this probability:

And we can find this probability with this difference:

d (5 x (-20)) * (100 x (0.7))

step-by-step explanation:

5÷1/100 * -.7 *-20

copy dot flip

5*100 * -.7 *-20

rearrange

5*-20   * 100 *-.7

-100   *-70

7000

i think that you should ask google for the answer to that problem because if we don't know it google does