The resting heart rate for an adult horse should average about μ = 42 beats per minute with a (95% of data) range from 18 to 66 beats per minute. Let x be a random variable that represents the resting heart rate for an adult horse. Assume that x has a distribution that is approximately normal.
(a) The empirical rule indicates that for a symmetrical and
bellshaped distribution, approximately 95% of the data lies within
two standard deviations of the mean. Therefore, a 95% range of data
values extending from μ – 2σ to μ +
2σ is often used for "commonly occurring" data values.
Note that the interval from μ – 2σ to μ
+ 2σ is 4σ in length. This leads to a "rule of
thumb" for estimating the standard deviation from a 95% range of
data values.Estimating the standard
deviation
For a symmetric, bellshaped distribution,
standard deviation ≈ 

≈ 

where it is estimated that about 95% of the commonly occurring
data values fall into this range.Use this "rule of thumb" to
estimate the standard deviation of x distribution. (Round
your answer to one decimal place.)
beats
(b) What is the probability that the heart rate is less than 25
beats per minute? (Round your answer to four decimal places.)
(c) What is the probability that the heart rate is greater than 60
beats per minute? (Round your answer to four decimal places.)
(d) What is the probability that the heart rate is between 25 and
60 beats per minute? (Round your answer to four decimal
places.)
(e) A horse whose resting heart rate is in the upper 12% 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 12%cutoff point of the probability
distribution? (Round your answer to the nearest whole
number.)
_______ beats per minute
a) standard deviation =range/4 =(6618)/4=12
b)
for normal distribution z score =(Xμ)/σx  
here mean= μ=  42 
std deviation =σ=  12.000 
probability that the heart rate is less than 25 beats per minute:
probability =  P(X<25)  =  P(Z<1.42)=  0.0788 
c)
probability that the heart rate is greater than 60 beats per minute:
probability =  P(X>60)  =  P(Z>1.5)=  1P(Z<1.5)=  10.9332=  0.0668 
d)
probability that the heart rate is between 25 and 60 beats per minute
probability =  P(25<X<60)  =  P(1.42<Z<1.5)=  0.93320.0788=  0.8554 
e)
for top 12% ; critical z =1.175
corresponding heart beat =mean+z*std deviaiton =56
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