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. significantly high (or at least 2 standard deviations above the mean). b. significantly low (or at least 2 standard deviations below the mean). c. not significant (or less than 2 standard deviations away from the mean).

Using the normal distribution, it is found that:

a) 2.28% of scores are significantly high.

b) 2.28% of scores are significantly low.

c) 95.44% of scores are not significant.

In a normal distribution with mean and standard deviation , the z-score of a measure X is given by:

Item a:

Z = 2 has a p-value of 0.9772.

1 - 0.9772 = 0.0228

Then, 2.28% of scores are significantly high.

Item b:

Z = -2 has a p-value of 0.0228.

Then, 2.28% of scores are significantly low.

Item c:

Z = 2 has a p-value of 0.9772.

Z = -2 has a p-value of 0.0228.

Then, 95.44% of scores are not significant.

A similar problem is given at link

set each factor equal to 0

4x-3 = 0

2x-1 = 0

solve for x.

4x-3=0

4x=3

x= 3/4

solve for x. 

2x-1= 0

2x= 1

x= 1/2

your answer should be 1/2 is less than or equal to x and 3/4 is greater than or equal to x

i saw this question before, dont know the asnwer srry

step-by-step explanation: