Complete Question
In a study on the physical activity of young? adults, pediatric researchers measured overall physical activity as the total number of registered movements? (counts) over a period of time and then computed the number of counts per minute? (cpm) for each subject. The study revealed that the overall physical activity of obese young adults has a mean of mu = 322 cpm and a standard deviation of 80 cpm
In a random sample of n=100 what is the probability that the mean overall physical activity level of the sample is between 300 and 310 cpm? P(300310) nothing
Answer:
The probability is [tex]P(300 < \= x < 310) = 0.06383[/tex]
Step-by-step explanation:
From the question we are told that
The mean is [tex]\mu = 322 \ cpm[/tex]
The standard deviation is [tex]\sigma = 80 \ cpm[/tex]
The sample size is [tex]n = 100[/tex]
Generally the standard error of the mean is mathematically represented as
[tex]\sigma_{\= x} = \frac{\sigma}{\sqrt{n} }[/tex]
=> [tex]\sigma_{\= x} = \frac{80}{\sqrt{100} }[/tex]
=> [tex]\sigma_{\= x} = 8[/tex]
Generally the probability that the mean overall physical activity level of the sample is between 300 and 310 cpm is mathematically represented as
[tex]P(300 < \= x < 310) = P(\frac{300 - 322}{8} < \frac{ \= x - \mu }{\sigma_{\= x}} < \frac{310 - 322}{8} )[/tex]
=> [tex]P(300 < \= x < 310) = P(-2.75 < \frac{ \= x - \mu }{\sigma_{\= x}} < -1.5 )[/tex]
Generally [tex] \frac{\= x -\mu}{\sigma } = Z (The \ standardized \ value\ of \ \= x ) [/tex]
=> [tex]P(300 < \= x < 310) = P(-2.75 < Z< -1.5 )[/tex]
=> [tex]P(300 < \= x < 310) = P(Z < -1.5) - P( Z< -2.75 )[/tex]
Generally the probability of (Z < -2.75) and ( Z< -1.5 )
[tex]P(Z < -2.75) = 0.0029798[/tex]
and
[tex]P(Z < -1.5 ) = 0.066807[/tex]
So
[tex]P(300 < \= x < 310) = P(Z < -1.5) - P( Z< -2.75 )[/tex]
=> [tex]P(300 < \= x < 310) = 0.066807 - 0.0029798[/tex]
=> [tex]P(300 < \= x < 310) = 0.066807 - 0.0029798[/tex]
=> [tex]P(300 < \= x < 310) = 0.06383[/tex]
