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What is the probability that the mean overall physical activity level of the sample is between 300 and 310​ cpm? ​P(300​310) nothing ​

User Bamieh
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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(300​310) nothing ​

Answer:

The probability is
P(300 < \= x < 310) = 0.06383

Explanation:

From the question we are told that

The mean is
\mu = 322 \ cpm

The standard deviation is
\sigma = 80 \ cpm

The sample size is
n = 100

Generally the standard error of the mean is mathematically represented as


\sigma_(\= x) = (\sigma)/(√(n) )

=>
\sigma_(\= x) = (80)/(√(100) )

=>
\sigma_(\= x) = 8

Generally the probability that the mean overall physical activity level of the sample is between 300 and 310​ cpm is mathematically represented as


P(300 < \= x < 310) = P((300 - 322)/(8) < ( \= x - \mu )/(\sigma_(\= x)) < (310 - 322)/(8) )

=>
P(300 < \= x < 310) = P(-2.75 < ( \= x - \mu )/(\sigma_(\= x)) < -1.5 )

Generally
(\= x -\mu)/(\sigma ) &nbsp;= &nbsp;Z (The &nbsp;\ standardized \ &nbsp;value\ &nbsp;of &nbsp;\ \= x )

=>
P(300 < \= x < 310) = P(-2.75 < Z< -1.5 )

=>
P(300 < \= x < 310) = P(Z < -1.5) - P( Z< -2.75 )

Generally the probability of (Z < -2.75) and ( Z< -1.5 )


P(Z < -2.75) = 0.0029798

and


P(Z < -1.5 ) = 0.066807

So


P(300 < \= x < 310) = P(Z < -1.5) - P( Z< -2.75 )

=>
P(300 < \= x < 310) = 0.066807 - 0.0029798

=>
P(300 < \= x < 310) = 0.066807 - 0.0029798

=>
P(300 < \= x < 310) = 0.06383

User Stephen Cleary
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