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Gala2k [10]
3 years ago
9

A data set with a mean of 34 and a standard deviation of 2.5 is normally distributed

Mathematics
1 answer:
tresset_1 [31]3 years ago
7 0

Answer:

a) z= \frac{34-34}{2.5}= 0

z= \frac{39-34}{2.5}= 2

And we want the probability from 0 to two deviations above the mean and we got 95/2 = 47.5 %

b) P(X

z= \frac{31.5-34}{2.5}= -1

So one deviation below the mean we have: (100-68)/2 = 16%

c) z= \frac{29-34}{2.5}= -2

z= \frac{36.5-34}{2.5}= 1

For this case below 2 deviation from the mean we have 2.5% and above 1 deviation from the mean we got 16% and then the percentage between -2 and 1 deviation above the mean we got: (100-16-2.5)% = 81.5%

Step-by-step explanation:

For this case we have a random variable with the following parameters:

X \sim N(\mu = 34, \sigma=2.5)

From the empirical rule we know that within one deviation from the mean we have 68% of the values, within two deviations we have 95% and within 3 deviations we have 99.7% of the data.

We want to find the following probability:

P(34 < X

We can find the number of deviation from the mean with the z score formula:

z= \frac{X -\mu}{\sigma}

And replacing we got

z= \frac{34-34}{2.5}= 0

z= \frac{39-34}{2.5}= 2

And we want the probability from 0 to two deviations above the mean and we got 95/2 = 47.5 %

For the second case:

P(X

z= \frac{31.5-34}{2.5}= -1

So one deviation below the mean we have: (100-68)/2 = 16%

For the third case:

P(29 < X

And replacing we got:

z= \frac{29-34}{2.5}= -2

z= \frac{36.5-34}{2.5}= 1

For this case below 2 deviation from the mean we have 2.5% and above 1 deviation from the mean we got 16% and then the percentage between -2 and 1 deviation above the mean we got: (100-16-2.5)% = 81.5%

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