PubHlth 540 Introductory Biostatistics Page 1 of 4 Unit 5 – Normal Distribution Practice Quiz SOLUTIONS 1. Suppose that the distribution of diastolic blood pressure in a population of hypertensive women is modeled well by a normal probability distribution with mean 100 mm Hg and standard deviation 14 mm Hg. Let X be the random variable representing this distribution. Find two symmetric values “a” and “b” such that probability [ a < X < b ] = .99 Answer: a=63.95 b=136.05 Solution: There is more than one approach for arriving at the same answer. Approach 1 is simpler. Approach 2 gives a better appreciation for the concepts involved Approach 1 – Simpler Step 1 – Launch the David Lane Normal distribution calculator that can be found on the course website page for topic 5. The Normal Distribution http://davidmlane.com/hyperstat/z_table.html Step 2 – Scroll down to the 2nd calculator that is provided. Enter 100 for the mean, 14 for the standard deviation, and 0.99 for the shaded area. Click on the button for “between”. The calculator returns the answer. … \web540 …quizsol_normal.docPubHlth 540 Introductory Biostatistics Page 2 of 4 Approach 2 – Shows detail of the formula used Step 1 – Identify symmetric values for the standard normal distribution such that the area enclosed is .99. Here, the idea is to recognize that the excluded area is .005 in each of the left and right tails. Thus, we want to find the 0.5th and the 99.5th percentiles. Launch the David Lane Normal distribution calculator that can be found on the course website page for topic 5. The Normal Distribution http://davidmlane.com/hyperstat/z_table.html Step 2 – Again, scroll down to the 2nd calculator that is provided. For the standard normal distribution, you should already see 0 for the mean and 1 for the standard deviation. Enter 0.99 for the shaded area. Click on the button for “between”. The calculator returns as the answer + 2.5758. … \web540 …quizsol_normal.docPubHlth 540 Introductory Biostatistics Page 3 of 4 Step 3 – Using the standardization formula as your starting point, solve backwards for the corresponding 0.5th and 99.5th percentiles of a normal distribution with mean 100 and standard deviation 14. x-μz = says that x=σ[z] + μσ Thus a = 0.5th percentile for X = 14[-2.57] + 100 = 63.95 and b = 99.5th percentile for X = 14[+2.57] + 100 = 136.05 2. Suppose that the distribution of weights of New Zealand hamsters is distributed normal with mean 63.5 g and standard deviation 12.2 g. If there are 1000 weights in this population, how many of them are 78 g or greater? Answer: 117 Solution: Pr [ weight > 78 g ] = Pr [ Normal μ=63.5 σ=12.2 > 78 ] 78-μ= Pr [ Standard normal > ] σ78-63.5= Pr [ Standard normal > ] 12.2= Pr [ Normal (0,1) > 1.1885 ] = .117 Therefore # Hamsters with weights > 78 g in a population of size 1000 = (Number of hamsters) x Pr[ weight > 78 g ] = (1000)(.117) = 117 … \web540 …quizsol_normal.docPubHlth 540 Introductory Biostatistics Page 4 of 4 3. Consider again the normal probability distribution of problem #2. What is the probability of selecting at random a sample of 10 hamsters that has a mean greater than 65 g? Answer: .3483 Solution: Tip – The solution to this problem requires noticing that the random variable is X , so that the standardization to Z must use the SE for this. n=10XX12.2Pr [ X > 65 g ] = Pr [ Normal μ =63.5 σ = > 65 ]10 XX65-μ65-63.5= Pr [ Standard normal > ] = Pr [ Standard normal > ] σ12.2 10 = Pr [ Normal (0,1) > 0.3888 ] = .3483 … \web540