Assume that females have pulse rates that are normally distributed with a mean of u = 77.0 beats per minute a standard deviation of o = 12.0 beats per minute.

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I am unable to figure out how the area (0.6915) in this example problem was reached. The problem says to "use technology", but doesn't specify how. I have tried to solve it multiple times in different ways using my TI-84 Plus calculator, but with no success. My calculator has given me errors several of the times that I have tried. Is there a step that I'm missing?

Assume that females have pulse rates that are normally distributed with a mean of u =77.0 beats per minute ar
a standard deviation of o = 12.0 beats per minute.
a. If 1 adult female is randomly selected, find the probability that her pulse rate is less than 83 beats per
minute.
Notice that an individual value from a normally distributed population has been chosen. Therefore, use the
population distribution to determine the probability.
First, convert the pulse rate to the corresponding z-score using z =
and the population distribution
statistics.
83 – 77.0
Z =
12.0
= 0.5
The area that corresponds to the probability that an adult female's pulse rate is less than 83 beats per minute
is the area under the standard normal distribution curve to the left of z = 0.5.
While technology or a standard normal distribution table can be
used to find the area to the left of z = 0.5, for this problem, use
taahnallau rai ndins ta faur dasimalnlann
Transcribed Image Text:Assume that females have pulse rates that are normally distributed with a mean of u =77.0 beats per minute ar a standard deviation of o = 12.0 beats per minute. a. If 1 adult female is randomly selected, find the probability that her pulse rate is less than 83 beats per minute. Notice that an individual value from a normally distributed population has been chosen. Therefore, use the population distribution to determine the probability. First, convert the pulse rate to the corresponding z-score using z = and the population distribution statistics. 83 – 77.0 Z = 12.0 = 0.5 The area that corresponds to the probability that an adult female's pulse rate is less than 83 beats per minute is the area under the standard normal distribution curve to the left of z = 0.5. While technology or a standard normal distribution table can be used to find the area to the left of z = 0.5, for this problem, use taahnallau rai ndins ta faur dasimalnlann
83 - 77.0
Z =
12.0
= 0.5
The area that corresponds to the probability that an adult female's pulse rate is less than 83 beats per minute
is the area under the standard normal distribution curve to the left of z = 0.5.
While technology or a standard normal distribution table can be
used to find the area to the left of z = 0.5, for this problem, use
technology, rounding to four decimal places.
The area is 0.6915.
X = 83
µ = 77.0
Press Continue to see more
Transcribed Image Text:83 - 77.0 Z = 12.0 = 0.5 The area that corresponds to the probability that an adult female's pulse rate is less than 83 beats per minute is the area under the standard normal distribution curve to the left of z = 0.5. While technology or a standard normal distribution table can be used to find the area to the left of z = 0.5, for this problem, use technology, rounding to four decimal places. The area is 0.6915. X = 83 µ = 77.0 Press Continue to see more
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