Question

Based on taking the pulses of most people in the United States, healthcare providers announce that 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. Find the probability that the average pulse of the 100 adults is between 78 and 81 beats per minute. Please round your answer to three decimal places!

Answer 1

Probability that the average pulse of the 100 adults is between 78 and 81 beats per minute is 0.7493.

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
`\bar X` = sample average pulse

The z score probability distribution for sample mean is given by;

Z =
`(\bar X-\mu)/((\sigma)/(√(n) ) )` ~ N(0,1)

where,
`\mu` = population mean pulse = 80 beats per minute

`\sigma` = 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 <
`\bar X` < 81)

P(78 <
`\bar X` < 81) = P(
`\bar X` < 81) - P(
`\bar X`
`\leq` 78)

P(
`\bar X` < 81) = P(
`(\bar X-\mu)/((\sigma)/(√(n) ) )` <
`(81-80)/((12)/(√(100) ) )` ) = P(Z < 0.83) = 0.79673

P(
`\bar X`
`\leq` 78) = P(
`(\bar X-\mu)/((\sigma)/(√(n) ) )`
`\leq`
`(78-80)/((12)/(√(100) ) )` ) = P(Z
`\leq` -1.67) = 1 - P(Z < 1.67)

= 1 - 0.95254 = 0.04746

Therefore, P(78 <
`\bar X` < 81) = 0.79673 - 0.04746 = 0.7493