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# CSUF CHEM 361B - Review on the normal distribution

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The Normal DistributionSlide 2Slide 3Solve the following problemsSolutionsSlide 6Slide 7Slide 8Slide 9Slide 10Sampling DistributionsSampling Distribution of the Sample MeanSlide 13Slide 14Slide 15Slide 16Solve the problemsSlide 18Sampling Distribution of the Sample Proportion (large samples)Slide 20Slide 21The Normal DistributionNotes and review problemsThe Normal Distribution•A normal variable is continuous, ranges from – to +, and forms a symmetrical distribution, with a mean  and a standard deviation The Normal Distribution•The density function f(x) is 2σμx21e2π1f(x)f(x)f(x)Solve the following problems1. Calculate the following probabilities:•P(0<Z<1.5)•P(Z<1.5)•P(-1.4<Z<.6)•P(Z>2.03)•P(Z>-1.44)•P(1.14<Z<2.43)•P(-.91<Z<-.31)2. Find Z0 •for which 4% of the population is located above it.•for which 2% of the population is located below it.•that satisfies the condition: P(-Z0<Z<+Z0) = .653. X is normally distributed with mean 100 and standard deviation 20. What is the probability that X is: 1. greater than 145? 2. Less than 1203. Less than 954. Between 75 and 100SolutionsSet 1:•P(0<Z<1.5) = P(-<Z<1.5) – 0.5 = .9332 – 0.5 = 0.4332With Excel use: =normsdist(1.5) – 0.5.•P(Z < 1.5) = P(-<Z<1.5) = 0.9332•P(-1.4 < Z < 0.6) = P(Z < 0.6) – P(Z < -1.4). Several steps are needed if using the Z table from the text: –P(Z < 0.6) = 0.7257–P(Z < -1.4) = P(Z > +1.4) = 1 – P(Z < 1.4) = 1 – 0.9192 = .0808–P(-1.4 < Z < 0.6) = 0.7257 – 0.0808 = .6449With Excel use: =normsdist(0.6) – normsdist(-1.4)•P(Z > 2.03) = 1 – P(Z < 2.03) = 1 – 0.9788 = .0212With Excel use: = 1 – normsdist(2.03)•P(Z > -1.44) = P(Z < 1.44) = 0.9251•P(1.14 < Z < 2.43) = P(Z < 2.43) – P(Z < 1.14) = 0.9925 – 0.8508 = .1417•P(-.91 < Z < -.31) = P(Z < -.31) – P(Z < -.91) = P(Z < +.91) – P(Z < +.31) = 0.8186 – 0.6217 = .1969With Excel use: =nornsdist(-.31) – normsdist(-.91).SolutionsSet 2:•P(Z > Z0) = .04. Thus, P(Z < Z0) = 0.96. The corresponding Z0= 1.75With Excel use: =normsinv(.96)•P(Z < Z0) = .02. The value of Z0 must be negative, therefore, if we want to use the text table we need to use symmetry. Instead of finding Z0 as requested, we’ll first find –Z0, which of course has a positive value! For convenience let’s call this value Z1P(Z > Z1) = .02. So, P(Z < Z1) = 0.98. Thus Z1 = 2.055 (or so). This means that the value we are seeking Z0 = -2.055.•P(-Z0 < Z < Z0) = 0.65. Then P(Z < Z0) – P(Z <- Z0) = 0.65. However since P(Z < -Z0) = P(Z > Z0) = 1 – P(Z < Z0), we have:P(-Z0 < Z < Z0) = P(Z < Z0) – [1 – P(Z < Z0)] = 2P(Z < Z0) – 1 = 0.65. From here we have P(Z < Z0) = 0.825, which yields Z0 =~0.935.Another solution: Note that P(Z<-Z0)+ P(-Z0 < Z < Z0) +P(Z>Z0)=1, but P(Z<-Z0)=P(Z>Z0) so, P(Z<-Z0)=.175, and –Z0 can be found.SolutionsSet 3:•P(X>145) = P(Z>(145 – 100)/20) = P(Z>2.25). Use the Z-table.With Excel use: =1 – normdist(145,100,20,True)•P(X<120) = P(Z<(120 – 100)/20) = P(Z<1). Use the table.With Excel use: =normdist(120,100,20,True)•P(X<95) = use the same approach as before•P(75<X<100) = P((75 – 100)/20<Z<(100 – 100)/20). Use the table.With Excel use: =normdist(100,100,20,True) – normdist(75,100,20,True)4. The lifetime of light-bulbs that are advertised to last for 5000 hours are normally distributed with a mean of 5,100 hours and a standard deviation of 200 hours.What is the probability that a bulb lasts longer than the advertised figure?If we wanted to be sure 98% of all the bulbs last longer than an advertised figure, what figure should be advertised? 5. Because of high interest rates, most consumers attempt to pay-off the credit card bills promptly. However this is not always possible. If the amount of interest paid monthly by card-holders is normally distributed with a mean of \$27 and a standard deviation of \$7,•What proportion of the cardholders pay more than \$30 in interest?•What proportion of the cardholders pay less than \$15 in interest?•What interest payment is exceeded by only 20% of the cardholders?Solve the following problemsSolutions4. Let X represent the bulb lifetime.•P(X>5100) = P(Z>(5000 – 5100)/200) = P(Z>-0.5) = P(Z<0.5)•Let the advertised figure be X0. P(X>X0) = .98; X0 must reside to the left of , therefore after transforming X into a Z variable, the latter must have a negative value. Let us call this Z-value –Z0 where –Z0 = (X0 – )/. Thus P(Z> –Z0) = .98 (click to see the graph), so P(Z<-Z0) = .02. From the table we have –Z0 = -2.055 (the mid-value between --2.05 whose probability is 0.0202, and 2.06 whose probability is .0197). So, the -Z0 = -2.055. From here we determine X0 by the equation -2.055 = (X0 – 5100)/200, or X0=5100 – 2.055(200) = 4689. The advertised lifetime should be “4689 hours” if it is desired that 98% of the bulbs will last at least that long.0.98-Z0Solutions5. Let X be the amount of interest paid monthly.•P(X>30) = P(Z>(30-27)/7) = P(Z>.42857). With Excel we have:= 1-normdist(30,27,7,true) =.334118•P(X<15) = P(Z<(15 –27)/7) = P(Z<-1.71429) = P(Z>1.71429) = 1-P(Z<1.71429)=1-normsdist(1.71429)=0.043238•Let X0 be the interest exceeded by 20% of the cardholders. P(X>X0) = 0.2; P(Z>Z0) = 0.2, where Z0 = (X0 – )/ Since Z0 is positive, to use the table we look at P(Z<Z0)=.80. The corresponding value of Z0 is 0.84. With Excel we use the function norminv or normsinv. These functions provide the X0 or the Z0 respectively for a left hand tail probability. Observe:X0: =norminv(.80,27,7) = 32.89; Z0 =normsinv(.80) = .8416 (to find X0 from Z0 note that X0 = +Z0 = 27 + .8416(7) = 32.89).0.20Z0Sampling DistributionsNotes and review problemsSampling Distribution of the Sample Mean•The sample mean is normally distributed if the parent distribution is normal.•Then sample mean is approximately normally distributed if the sample is sufficiently large (n30) even if the parent distribution is not normal. The larger the sample the better the approximation.•The mean of the sample mean is the same as the parent population mean•The standard deviation of the sample mean is )μμ(xxnσσxxSolve the following problems1. A sample of n=16 observations is drawn from a normal population with =1,000 and =200. Find the following probabilities:2. The heights of North American women are normally distributed with a mean of 64 inches and a

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