Assuming a normal distribution of 1000 cases, how many cases will be farther away from the mean than + 3 standard deviations? a. about 3 b. at least 500 c. it is impossible to estimate d.327

And then the probability that  will be farther away from the mean than + 3 standard deviations is just

1-0.9973= 0.0027

And the number of cases would be 1000*0.0027=2.7 or approximatley 3 cases. And the bes option is : a.About 3

Step-by-step explanation:

1) 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".  

2) Solution to the problem

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

We are interested on this probability

And the best way to solve this problem is using the normal standard distribution and the z score given by:

And we can find this probability on this way:

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.  

And then the probability that  will be farther away from the mean than + 3 standard deviations is just

1-0.9973= 0.0027

And the number of cases would be 1000*0.0027=2.7 or approximatley 3 cases. And the bes option is : a.About 3

29 mil

step-by-step explanation:

it's equivalent, so we can say that its equal.

let's call the denominator x and the numerador is 18 less then denominator, x - 18.

multiply

2.x = 5.(x - 18)

2x = 5x - 90

90 = 5x - 2x

90 = 3x

x = 90/3

x = 30

so: