# When a mean is 40 and standard dev. is 7.5, how do you find probability that a given value will be greater than 55.75?

#P_(x>55.75)=0.0179#

Given -

Mean

SD

At

#z=(x-mu)/sigma=(55.75-40)/7.5=15.75/7.5=2.1#

Probability that a given value will be greater than 55.75 = [Area between z=0 and z=

#P_(x>55.75)=0.5-0.4821=0.0179#

#P_(x>55.75)=0.0179#

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To find the probability that a given value will be greater than 55.75, you would first calculate the z-score using the formula ( z = \frac{x - \mu}{\sigma} ), where ( x ) is the given value, ( \mu ) is the mean, and ( \sigma ) is the standard deviation. Then, you would use a standard normal distribution table or a calculator to find the probability corresponding to that z-score.

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To find the probability that a given value will be greater than ( 55.75 ) in a normal distribution with a mean of ( 40 ) and a standard deviation of ( 7.5 ), you can use the z-score formula and then look up the corresponding area under the standard normal curve.

The z-score formula is:

[ z = \frac{x - \mu}{\sigma} ]

Where:

- ( x ) is the given value (( 55.75 ) in this case)
- ( \mu ) is the mean (( 40 ))
- ( \sigma ) is the standard deviation (( 7.5 ))

Calculate the z-score:

[ z = \frac{55.75 - 40}{7.5} ] [ z = \frac{15.75}{7.5} ] [ z = 2.1 ]

Now, you can look up the probability corresponding to a z-score of ( 2.1 ) in a standard normal distribution table or using a calculator. The probability of a z-score greater than ( 2.1 ) is the same as the probability of a value greater than ( 55.75 ) in the original distribution.

From the standard normal distribution table or calculator, the probability corresponding to ( z = 2.1 ) is approximately ( 0.0179 ).

So, the probability that a given value will be greater than ( 55.75 ) is approximately ( 0.0179 ), or ( 1.79% ).

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To find the probability that a given value will be greater than 55.75, you can use the z-score formula and the standard normal distribution table.

First, calculate the z-score of 55.75 using the formula:

[ z = \frac{x - \mu}{\sigma} ]

where ( x ) is the given value (55.75), ( \mu ) is the mean (40), and ( \sigma ) is the standard deviation (7.5).

[ z = \frac{55.75 - 40}{7.5} ] [ z = \frac{15.75}{7.5} ] [ z \approx 2.1 ]

Next, use the standard normal distribution table or a calculator to find the probability associated with a z-score of approximately 2.1. This represents the probability that a value randomly selected from the distribution will be greater than 55.75.

From the standard normal distribution table, the probability corresponding to a z-score of 2.1 is approximately 0.0179.

So, the probability that a given value will be greater than 55.75 is approximately 0.0179, or 1.79%.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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