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5.4- Pulse Rates of Women. Women have pulse rates that are normally distributed

ID: 3327797 • Letter: 5

Question


5.4- Pulse Rates of Women. Women have pulse rates that are normally distributed with a mean of 77.5 beats per minute and a standard deviation of 11.6 beats per minute (based on the data in the Lab 5 tables). a) Find the percentiles Pi and P99. TI83/84: You are given the mean and standard deviation. You need to calculate the total area to the left of the desired z score for both Pi and P99 Step 1: Press "2nd” and then “DISTR" Step 2: Select “inv norm( Step 3: The three values must be entered in the correct sequence (separated by commas). a) the area to the left of the desired z score b) mean c) standard deviation Step 4: Press “enter" to get the results. b) Dr. Puretz sees exactly 25 female patients each day. Find the probability that 25 randomly selected women have a mean pulse rate between 70 beats per minute and 85 beats per minute. Using the equation Ox-C , Find the standard deviation of sampling distributions of means Using the mean bpm from and the standard deviation you calculated, find the probability between x = 70 bpm and x = 85 bpm. TI83/84: Step 1: Press “2nd” and then “DISTR Step 2: Select ".normalcdf Step 3: The four values must be entered in the correct sequence (separated by commas). a) the lower boundary b) the upper boundary c) mean d) standard deviation Step 4: Press "enter" to get the results. c) If Dr. Puretz wants to select pulse rates to be used as cutoff values for determining when further tests should be required, which pulse rates are better to use: the results from part (a) or the pulse rates of 70 beats per minute and 85 beats per minute from part (b)? Why? For this last part read the question carefully & explain your answer

Explanation / Answer

a) for P1 and P99 ; z score =-/+ 2.326

hence P1=mean +z*std deviation =77.5-2.326*11.6 =50.51

and P99=mean +z*std deviation =77.5+2.326*11.6 =104.49

b) std error of mean =std deviation/(n)1/2 =11.6/(25)1/2 =2.32

hence probability =P(70<X<85)=P((70-77.5)/2.32 <Z<(85-77.5)/2.32) =P(-3.2328<Z<3.2328)=0.9994-0.0006

=0.9988

c)part A values are required as it gives better picture for an individual; part B is for mean values of 25 female which always indicate values closer to the mean

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