Answer:
0.0015 is he probability that a randomly selected student has a body weight of greater than 169 pounds.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 142 pounds
Standard Deviation, σ = 9 pounds
We are given that the distribution of body weights is a bell shaped distribution that is a normal distribution.
Empirical Rule:
- The empirical rule states that for a normal distribution 68% falls within the first standard deviation from the mean, 95% within the first two standard deviations from the mean and 99.7% within three standard deviations of the mean.
P( body weight of greater than 169 pounds)
![169 = 142 + 3(9) = \mu + 3\sigma\\115 = 142-3(9) = \mu - 3\sigma](https://tex.z-dn.net/?f=169%20%3D%20142%20%2B%203%289%29%20%3D%20%5Cmu%20%2B%203%5Csigma%5C%5C115%20%3D%20142-3%289%29%20%3D%20%5Cmu%20-%203%5Csigma)
According to empirical rule, 99.7% within three standard deviations of the mean.
Thus, we can write:
P( body weight of greater than 169 pounds)
![\displaystyle\frac{1-P(\text{Body weight between 115 and 169})}{2}\\\\= \frac{1-0.997}{2} = 0.0015](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cfrac%7B1-P%28%5Ctext%7BBody%20weight%20between%20115%20and%20169%7D%29%7D%7B2%7D%5C%5C%5C%5C%3D%20%5Cfrac%7B1-0.997%7D%7B2%7D%20%3D%200.0015)
0.0015 is he probability that a randomly selected student has a body weight of greater than 169 pounds.
Answer:
540
Step-by-step explanation:
9 times 60 equals 540
Answer:
-30
Step-by-step explanation:
If you add -30 to both sides of the equation, you get 15 = m.
Answer:
18 ducks and 9 pigs i believe...
Step-by-step explanation: