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Natali [406]
3 years ago
7

For bone density scores that are normally distributed with a mean of 0 and a standard deviation of 1, find the percentage of sco

res 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.
Mathematics
1 answer:
Alex787 [66]3 years ago
6 0

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

b) P(Z

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

P(Z

c) P(-2

And we can find this probability with this difference:

P(-2

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

Part b

We are interested on this probability

P(Z

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

P(Z

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

For this case we want this probability:

P(-2

And we can find this probability with this difference:

P(-2

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