Respuesta :
Answer:
Probability that the average pulse of the 100 adults is between 78 and 81 beats per minute is 0.7493.
Step-by-step explanation:
We are given that adult between the age of 21 and 65 have a mean pulse of 80 beats per minute with a standard deviation of 12 beats per minute.
A doctor measures 100 random, independent adults between the ages of 21 and 65 from the population of such adults and calculates their average pulse.
Let [tex]\bar X[/tex] = sample average pulse
The z score probability distribution for sample mean is given by;
Z = [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = population mean pulse = 80 beats per minute
[tex]\sigma[/tex] = standard deviation = 12 beats per minute
n = sample of adults measured = 100
Now, probability that the average pulse of the 100 adults is between 78 and 81 beats per minute is given by = P(78 < [tex]\bar X[/tex] < 81)
P(78 < [tex]\bar X[/tex] < 81) = P([tex]\bar X[/tex] < 81) - P([tex]\bar X[/tex] [tex]\leq[/tex] 78)
P([tex]\bar X[/tex] < 81) = P( [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\frac{81-80}{\frac{12}{\sqrt{100} } }[/tex] ) = P(Z < 0.83) = 0.79673
P([tex]\bar X[/tex] [tex]\leq[/tex] 78) = P( [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] [tex]\leq[/tex] [tex]\frac{78-80}{\frac{12}{\sqrt{100} } }[/tex] ) = P(Z [tex]\leq[/tex] -1.67) = 1 - P(Z < 1.67)
= 1 - 0.95254 = 0.04746
Therefore, P(78 < [tex]\bar X[/tex] < 81) = 0.79673 - 0.04746 = 0.7493
