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Answer :
Answer:
mean=142
169-142/9
P(x>169)=1-.997/2
=.15
Step-by-step explanation:
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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)
[tex]169 = 142 + 3(9) = \mu + 3\sigma\\115 = 142-3(9) = \mu - 3\sigma[/tex]
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)
[tex]\displaystyle\frac{1-P(\text{Body weight between 115 and 169})}{2}\\\\= \frac{1-0.997}{2} = 0.0015[/tex]
0.0015 is he probability that a randomly selected student has a body weight of greater than 169 pounds.
