Department of Civil Engineering-I.I.T. Delhi CEL 899: Environmental Risk Assessment Statistics and Probability Example Part 1
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1 Department of Civil Engineering-I.I.T. Delhi CEL 899: Environmental Risk Assessment Statistics and Probability Example Part Note: Assume missing data (if any) and mention the same. Q. Suppose X has a normal distribution defined as N (mean=5, variance= ) (note the notation used here), answer the following (i) Calculate mean, median, standard deviation, variance (ii) Calculate P(X=3), P(X<0)? (iii) (iv) Calculate P (<X<8), i.e., probability of X obtaining between and 8? Hint: P(X<X<X) = P(X<X)-P(X<X)? (v) Calculate 5 th, 50 th, 90 th, 95 th, and 99 th percentile values. Also calculate 90% confidence interval value (= 95 th percentile value 5 th percentile)? Answer: (i) Mean = 5; Standard deviation =; Variance = = 4 (Answer) At median value, 50% of X values are lower than this value and 50% of X values are higher than this. So, P(X< Median) = P(X> Median) =0.50 So if we determine 50 th percentile value, it will be median value for random variable X with N (mean=5, variance= ). Given that P(X<median) = 0.5 or P([(X-5)/]< [(median-5)/]) =0.5 here Z=[(X-5)/] and Z * =[(median-5)/] So we have P(Z<Z*)=0.5 where Z has a standard normal distribution with mean 0 and standard deviation equal to. Now look at your standard normal distribution table to find Ф (Z * )= P(Z<Z * ) = 0.5. The table gives Z * = 0 i.e., Z * =[(median-5)/]=0 => median = 5 (Answer) Note: For normal distribution, mean = median. But I showed you this approach for calculating any percentile value. Here we did the calculation for 50 th percentile value (i.e, median). Similarly, you can calculate all 5 th, 50 th, 95 th percentile values. (ii) P(X=3) =? Put X=3 in the following formula of probability function for normal distribution. X µ f ( x, µ, σ ) = exp 0.5 σ π σ => 3 5 f (3;5,) = exp 0.5 π => P(X=3) = f(3; 5,) = 0. (Answer) = (/5.0) * exp(-0.5) = (/5.0)*65 Now, P(X<0)=? First get Z value which has a standard normal distribution with mean 0 and standard deviation equal to. or P([(X-5)/]< [(0-5)/]) =? here Z=[(X-5)/] and Z * =[(0-5)/] =.5 So we have P(Z<.5)=?
2 Now look at your standard normal distribution table to find Ф (.5) as Z*=.5 is calculated and known here. => P(Z<Z * ) = So, P(X<0) = P(Z<Z * ) = (Answer) (iii) P(<X<8) = P(X<8)-P(X<) First get Z value which has a standard normal distribution with mean 0 and standard deviation equal to. P(X<8)= P([(X-5)/]< [(8-5)/]) = P(Z <.5] = Ф (.5) =0.933 P(X<)= P([(X-5)/]< [(-5)/]) = P(Z < -.0] = Ф (-.0) (or P(Z<-)) Here, remember the formula: P(Z<Z * )=P(Z>-Z*) (because of symmetry of normal distribution about mean). so, P(Z<-) = P(Z>)=-P(Z<) So, Ф (-.0) =- Ф (.0) = = 0.07 So, P(<X<8) = P(X<8)-P(X<) = = => (i.e., 0.905*00 = 9.05% of times X will lie between and 8). (answer) Note: Here 9.05% confidence interval is given by (, 8). (iv) Calculate all percentile values as we calculated for 50 th percentile values in part (i). Say 5 th percentile value = X* i.e., P(X<X * )=0.05 P(X<X*)= P([(X-5)/]< [(X*-5)/]) = P(Z < Z*) = 0.05 [here, Z*= (X*-5)/] As P (Z<Z*) = -P(Z>Z*) 0.05=-P(Z>Z*) P(Z>Z*)=0.95 Now, as P(Z<Z * )=P(Z>-Z*) (because of symmetry of normal distribution about mean), it means that P(Z>Z*)= P(Z<-Z*). As we know from table that for P(Z>Z*)=0.95= P(Z<-Z*) Here, Ф (Z)= 0.95 happens for Z lying between.64 and.65. So using linear interpolation, Z comes out to be.645. So, it means that Z* =.645 Z*=-.645 Now as Z*= (X*-5)/ = X* = *(-.645)+5 =.7 So 5 th percentile value =.7 Q. Look at the following failure data: Failure no Operating time (days) (say X) Using the operating time data, answer the following (i) Calculate mean, median, standard deviation, variance, minimum, maximum operating time values (ii) Develop a frequency histogram for operating time before failure (random variable X) using the binning approach (called as probability mass function, if X is discrete or probability density function if X is continuous). Develop this histogram for both cases. (iii) Develop cumulative distribution functions for two cases: For X as a discrete random variable) and for X as a continuous variable. Plot. (iv) Using the cumulative distribution function for X as a discrete, calculate P(X=0 days), P(X< 0 days). Repeat this calculation for the case when X is a continuous variable. (v) Calculate operating time without failure with 95 % confidence (i.e., calculate 95 th percentile value) using the developed cumulative density function when X is a continuous variable case. Also calculate 90% confidence interval value (= 95 th percentile value 5 th percentile)? Note: 90% confidence interval value indicates that your operating time lies between these two ends 90% of the observation times.
3 Answer: (i) First arrange the data in increasing order. Arranged data: X (in days): 40, 98, 65, 35, 3, 48, 547, 70, 95, 340 (total N=0) Minimum = 40 days, maximum = 340 days Mean = N = 0 X = x i N = 0 =(/0) *[ ] i= => mean = 48 days As N is even number here, median = average of two middle terms = ½ (5 th term + 6 th term) Median =/*(3+48)=370 days Standard deviation (σ) σ = σ = N = 0 i= 0 ( X X ) i N [( 40 48) + ( 98 48) +...( ) ] Table. Operating time (days) (say X) (X-mean) Total sum of different (X-mean) terms = So, standard deviation (σ) = (59406/9) 0.5 = (69934) 0.5 = 4. days Variance = σ = (4.) = days (ii) Table. Failure no. operating frequency (X/total sum) F(X)= P(X<x) time (days) [i.e. ) 40 40/480 40/ /480 38/ / / / / / / /480 78/ /480 85/ / / / / /480.0 total sum 480 3
4 .00 Probability (or frequency) histogram Fig. Probability (or frequency) histogram (see attached spreadsheet) (X=discrete; thus points are not connected here) Probability (or frequency) histogram (X=continuous) Fig. Probability (or frequency) plot (see attached spreadsheet) (X=continuous; thus points are connected here) (iii).00 Cumulative Probability Histogram Fig.3 Cumulative Probability Histogram (see attached spreadsheet) (X=discrete; thus points are not connected here) 4
5 Cumulative Probability Histogram (X=continuous) Fig.4 Cumulative Probability Plot (see attached spreadsheet) (X=continuous; thus points are connected here) (iv) For X: discrete variable. P(X=0 days)=? Refer Table. As X is discrete and as per the data, it does not have any value at X=0, so P( X=0 days) =0 (answer) Now, P(X< 0 days)= P(X=98) +P(X=40) (as X is smaller than 0 days) = (98/480)+(40/480)=38/480 (answer) For X: continuous variable. P(X=0 days)=? Refer Table. As X is continuous variable, it can take any value including X=0. So X=0 lies between 98 days and 65 days. Do linear interpolation to determine frequency of getting X=0 days (or P(X=0 days)): (0-98)/(65-98) = (P(X)-98/480)/(65/480-98/480)=> so, P(X=0) = 0/480 (answer) Now, P(X< 0 days)=? Refer Table. As X is continuous variable, assume linearity holds for F(x) function as well (see Figure 4). So F(X<=0) lies between F(X<=65) and F(X<=98). Do linear interpolation to determine cumulative probability of getting operating time < = 0 days: (0-98)/(65-98) = (F(0)-38/480)/(303/480-38/480) ()/(67) = (F(0)-38/480)/(65/480) => (/67) *(65/480) = F-(38/480) => F (X=0)= P(X<=0) = 9.5/480 (answer) (v) From table : Table 3. Failure no. operating time (days) F(X)= P(X<x) 40 40/480 = /480 = /480= /480= /480= /480= /480= /480= /480= =.0 5
6 Assuming X as a continuous variable and its cumulative probability density function is given by F(X) as per Table 3, first determine 5 th percentile value and 95 th percentile and then calculate 90% confidence interval. 5 th percentile value = that X * for which F(X * )=0.05 From table 3, it lies between F(98) =0.087 and F(65)= Say after linear interpolation, 0.05 value of F(X*) comes for X * = 30 days (assumed for illustration; calculate in homework and exam). Similarly, F(X**) =0.95 comes out to be 00 days (say) (again assumed; you should calculate it). So 90% confidence interval value for operating time before failure (days) = 95 th percentile value -5 th percentile value =00 days-30 days = 070 days (answer) 90% confidence range: (5 th percentile value, 95 th percentile value) = (30 days, 00 days) 6
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