Which of the following statements are true? Please help

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

tresset_1 [31]

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%

7 0

3 years ago

Solve for x <br> 8x+1>−31

slavikrds [6]

$8x + 1 > - 31 \\ 8x > - 1 - 31 \\ 8x > - 32 \\ x > \frac{ - 32}{8} \\ x > - 4$