For a normal distribution, is it likely that a data value selected at random is more than 2 standard deviations above the mean? No. This event happens only 2.5% of the time. No. This event happens only 16% of the time. No. This event happens only .015% of the time. Yes. This event happens quite frequently.

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

Using the z score formula we want this probability:

And using the complement rule we got:

And that correspond to only 0.023*100= 2.3%, so is not frequently and the most appropiate answer for this case would be:

No. This event happens only 2.5% of the time

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".  

Solution to the problem

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

Where the mean and  the deviation

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:

Using the z score formula we want this probability:

And using the complement rule we got:

And that correspond to only 0.023*100= 2.3%, so is not frequently and the most appropiate answer for this case would be:

No. This event happens only 2.5% of the time

find the magnitude of the trig term  

csc

(

x

)

−

1

by taking the absolute value of the coefficient.

1

find the lower bound of the range.

tap for more

y

=

−

2

find the upper bound of the range.

tap for more

y

=

0

the range is  

y

≤

−

2

or  

y

≥

0

.

(

−

∞

,

−

2

]

∪

[

0

,

∞

)

{

y

|

y

≤

−

2

,

y

≥

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}

step-by-step explanation: