Hi Makena! I'm assuming your teacher wants you to use z statistics to determine these probabilities. We'll be using the equation z = (x - μ) / σ.
However, before we can answer most of these questions we first need to calculate the standard deviation of our population. We know that the mean of the population is μ = 47 beats and that 95% of data lies between 24 to 70 beats per minute. This means that on our normal distribution, 2.5% of data is less than 24 beats and 97.5% of data is less than 70 beats. Using the middle part of a z-table or whatever z-score/percentile calculator you prefer, you'll see that 95% of data is approximately 1.96 standard deviations away from the mean (that is, z = 1.96). Rearranging our original equation a bit to isolate the standard deviation:
σ = (x - μ)/z = (70 - 47)/1.96 = 11.7347 = 12 (I'm going to round to 12, since heart beats really only occur in whole numbers)*
*You can also use x = 24 rather than x = 70, as long as you remember that 24 is NEGATIVE 1.96 standard deviations away from the mean. σ = (x - μ)/z = (24 - 47) / -1.96 = 11.7347.
b. What is the probability that the heart rate is less than 25 beats per minute?
z = (25 - 47) / 12
z = -1.83
Using a z-table or calculator, z = -1.83 gives a probability of 0.0336. This means P( x< 25) = 0.0336= 03.36%.
c. What is the probability that the heart rate is greater than 60 beats per minute? (Round your answer to four decimal places.)
z = (60 - 47) / 12
z = 1.08
z = 1.08 gives a probability of 0.8599. However, this question wants you to know what the probability of obtaining GREATER than 60 beats per minute, and most z tables tell you the probability of getting LESS than 60 beats per minute. Therefore, we must subtract our value from 1.
P( heart rate > 60) = 1 - 0.8599 = 0.1401 = 14.01%
(d) What is the probability that the heart rate is between 25 and 60 beats per minute? (Round your answer to four decimal places.)
Don't need the z-table for this one, we already did the calculations! If P(x<25) = 0.0336 and P(x>60) = 0.1401, then our answer must be whatever probability is left over.
P( 25<x<60) = 1 - (0.0336 + 0.1401) = 0.8266
(e) A horse whose resting heart rate is in the upper 9% of the probability distribution of heart rates may have a secondary infection or illness that needs to be treated. What is the heart rate corresponding to the upper 9% cutoff point of the probability distribution? (Round your answer to the nearest whole number.)
beats per minute
If the horse is in the upper 9%, then that means 91% of horses have heart beats below it. 0.91 gives a z-value of approximately 1.34.
z = (x - μ)/σ
x = z * σ + μ
x = 1.34 * 11 + 47
x = 61.74 = 62 beats per minute
