According with the 68–95–99.7 rule for the Normal distribution: If $\large \bar x$ is the mean of the distribution and s the standard deviation, around 68% of the data must fall in the interval

$\large [\bar x - s, \bar x +s]$

around 95% of the data must fall in the interval

$\large [\bar x -2s, \bar x +2s]$

around 99.7% of the data must fall in the interval

$\large [\bar x -3s, \bar x +3s]$

So, the range of lengths that covers almost all the data (99.7%) is the interval

[39.2 - 3*2.2, 39.2 + 3*2.2] = [32.6, 45.8]

This means that if we measure the upper arm length of a male over 20 years old in the United States, the probability that the length is between 32.6 cm and 45.8 cm is 99.7%

3 0

3 years ago

Solve for p P-17=2p+3

Y_Kistochka [10]

Answer:

-20

Step-by-step explanation:

P-17=2P+3

P-2P=17+3

-P=20

Divide both sides by negative 1

P=20

7 0

4 years ago