Alright, let's break down the given information and understand what conclusions we can draw about your teacher's pulse rate.
Firstly, the data on pulse rates are normally distributed, which means it follows the bell curve. We are given:
- The mean (average) pulse rate of the population, which is 72 beats per minute.
- The standard deviation, which measures how spread out the numbers are from the mean, is 3.5 beats per minute.
- Your teacher's pulse rate is 65 beats per minute.
To find out where your teacher's pulse rate falls in the distribution, we use the z-score formula:
[tex]\[ z = \frac{(X - \mu)}{\sigma} \][/tex]
Where:
- [tex]\( X \)[/tex] is your teacher’s pulse rate (65 bpm in this case),
- [tex]\( \mu \)[/tex] is the population mean (72 bpm),
- [tex]\( \sigma \)[/tex] is the standard deviation (3.5 bpm).
Plugging the values into the formula gives:
[tex]\[ z = \frac{(65 - 72)}{3.5} = \frac{-7}{3.5} = -2.0 \][/tex]
The z-score of -2.0 tells us how many standard deviations your teacher’s pulse is below the mean.
Next, we need to determine the percentile, which tells us the relative standing of your teacher's pulse rate within the distribution. The z-score of -2.0 corresponds to a percentile which gives the probability that a score is below this value in a standard normal distribution.
Looking up the z-score of -2.0 in the cumulative distribution function (CDF) table or using a CDF calculator, we find that the percentile is approximately 2.28%. This means that approximately 2.28% of the population has a pulse rate that is equal to or below your teacher's pulse rate.
Therefore, the correct conclusion is:
2.5% of the population has a pulse rate equal to or below your teacher's pulse rate because it is at the 2.5th percentile.
Given the answer choices, the closest match is the first option, even though the exact percentile is 2.28%, for practical purposes, it is reasonable to approximate it to 2.5%.
