Question

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.

Answer 1

Answer:

a) $P(Z>2)$

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

$P(Z>2) =1-P(Z<2) = 1-0.97725= 0.0228$

b) $P(Z<-2)$

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

$P(Z<-2) = 0.0228$

c) $P(-2<Z<2)$

And we can find this probability with this difference:

$P(-2<Z<2)=P(Z<2)-P(Z<-2) = 0.97725-0.02275= 0.9545$

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:

$Z \sim N(0,1)$  

Where $\mu=0$ and $\sigma=1$

We are interested on this probability

$P(Z>2)$

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

$P(Z>2) =1-P(Z<2) = 1-0.97725= 0.0228$

Part b

We are interested on this probability

$P(Z<-2)$

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

$P(Z<-2) = 0.0228$

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

For this case we want this probability:

$P(-2<Z<2)$

And we can find this probability with this difference:

$P(-2<Z<2)=P(Z<2)-P(Z<-2) = 0.97725-0.02275= 0.9545$