Solution Found!
For bone density scores that are normally distributed with
Chapter 6, Problem 49BSC(choose chapter or problem)
Problem 49BB
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 are
a. within 1 standard deviation of the mean.
b. more than 2 standard deviations away from the mean.
c. within 1.96 standard deviations of the mean.
d. between μ – 2σ and μ + 2σ.
e. more than 3 standard deviations away from the mean.
Questions & Answers
QUESTION:
Problem 49BB
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 are
a. within 1 standard deviation of the mean.
b. more than 2 standard deviations away from the mean.
c. within 1.96 standard deviations of the mean.
d. between μ – 2σ and μ + 2σ.
e. more than 3 standard deviations away from the mean.
ANSWER:Answer:
Step 1:
Given that,for bone density scores that are normally distributed with a mean of 0 and a standard deviation of 1.
a). The percentage of scores that are within 1 standard deviation of the mean is:
We are looking for a probability of scores that are within 1 standard deviation of the mean so we would need to find the z-standard scores for both numbers.
Therefore,
Z =
=
= 1.
Z =
=
= -1.
Now we need to subtract the probability of scores that are within 1 standard deviation of mean +1 and -1 is
P(a < Z < b) = P(Z < b) - P(Z < a)
P(-1 < Z < 1) = P(Z < 1.00) - P(Z < -1.00)
= 0.8643 - 0.1587
= 0.6826.
We can conclude that the probability of scores that are within 1 standard deviation of the mean is 0.6826. But we need to find the percentage the probability of scores that are within 1 standard deviation of the mean. So we need to multiply area of the curve with 100. We get nearly 68.26% the probability of scores that are within 1 standard deviation of mean.