Questions and Answers #1 Normal Distributions

Questions and Answers Sheet 1 Normal Distributions Question #1 Describe in words the surface whose equation is given. (assume that r is not negative.) 𝜃 = a) The plane b) The plane c) The plane d) The plane e) The plane 𝜋 4 y = -z where y is not negative y = z where y and z are not negative y = x where x and y are not negative y = -x where y is not negative x = z where x and y are not negative Answer: It represents line equation in XY- plane whose equation if given by x=y where x and y are not negative Question #2 Use the following Normal Distribution table to calculate the area under the Normal Curve (Shaded area in the Figure) when Z=1.3 and H=0.05; Assume that you do not have vales of the area beyond z=1.2 in the table; i.e. you may need to use the extrapolation. Check your calculated value and compare with the values in the table [for z=1.3 and H=0.05] . Calculate your percentage of error in the estimation. How do I solve this problem using extrapolation? Z+H 1.20000 1.21000 1.22000 1.23000 1.24000 1.25000 1.26000 1.27000 1.28000 1.29000 1.30000 1.31000 1.32000 1.33000 1.34000 1.35000 Prob. Extrapolation 0.38490 Differences 0.38690 0.00200 0.38880 0.00190 0.39070 0.00190 0.39250 0.00180 0.39440 0.00190 0.39620 0.00180 0.39800 0.00180 0.39970 0.00170 0.40150 0.00180 0.40320 0.00170 0.40490 0.00170 0.40660 0.00170 0.40830 0.00170 0.41010 0.00180 0.41190 0.00180 Answer: Here, we have to use the technique of extrapolation for finding the probability for the z = 1.35 (Z=1.3 & H = 0.05). Let us see this discuss this question step by step given as below: For using extrapolation we need the values for the row of z = 1.2 The values for the row z and columns H are given as below: By using the extrapolation, we get Z+H 1.20000 1.21000 1.22000 1.23000 1.24000 1.25000 1.26000 1.27000 1.28000 1.29000 1.30000 1.31000 1.32000 1.33000 1.34000 1.35000 Prob. Extrapolation 0.38490 Differences 0.38690 0.00200 0.38880 0.00190 0.39070 0.00190 0.39250 0.00180 0.39440 0.00190 0.39620 0.00180 0.39800 0.00180 0.39970 0.00170 0.40150 0.00180 0.40320 0.00170 0.40490 0.00170 0.40660 0.00170 0.40830 0.00170 0.41010 0.00180 0.41190 0.00180 We get P(Z<1.35) = 0.4119 by using the extrapolation however the probability by using the table is given as 0.4115. This means there is a very small difference observed due extrapolation. Question #3 Statistics from Cornell's Northeast Regional Climate Center indicate that Ithaca, New York, gets an average of 35.4'' of rain each year, with a standard deviation of 4.2''. Assume that a Normal model applies. a) During what percentage of years does Ithaca get more than 40'' of rain? b. Less than how much rain falls in the driest 20% of all years? c. A Cornell University student is in Ithaca for 4 years. Let 𝑦̅ represent the mean amount of rain for those 4 years. Describe the sampling distribution model of this sample mean, 𝑦̅ . d) What 's the probability that those 4 years average less than 30'' of rain? Answer: Step 1 (a) 𝜇 = 35.4 𝜎 = 4.2 40 − 35.4 𝑃(40 ≤ 𝑋) = 𝑃 ( ≤ 𝑧) = 𝑃(1.095 ≤ 𝑧) 4.2 𝑃(1.095 ≤ 𝑧) = 0.1368 Hence during 13.68% percentage of years does Ithaca get more than 40'' of rain. Step 2 (𝑏) 𝑃(𝑍 ≤ 𝑧) = 0.2 By Z-table: z = −0.84 𝑥−35.4 4.2 = −0.84 𝑥 = 31.87 Hence the driest 20% of days there will be approximately 31.87'' of rain. Step 3 (c) The shape of the sampling distribution model of the sample mean will be normal distribution. 𝜇 = 35.4 4.2 𝜎= = 2.1 √4 Step 4 (d) 𝑃(𝑋 < 30) = 𝑃 (𝑧 < 30−35.4 2.1 ) = 𝑃(𝑧 < −2.57) = 0.00506 Therefore, probability that those 4 year average less than 30'' of rain is 0.00506 Result (a) 13.68% (b) 31.87'' (c) Normal, Mean 35.4, Standard deviation 2.1 (d) 0.00506 Question #4 1) If A and B are mutually exclusive events with P(A) = 0.3 and P(B) = 0.5, then 𝑃(𝐴 ∩ 𝐵) =? 2) An experiment consists of four outcomes with 𝑃(𝐸1 ) = 0.2, 𝑃(𝐸2 ) = 0.3, and 𝑃(𝐸3 ) = 0.4.The probability of outcome E_4 is ? 3) If A and B are mutually exclusive events with P(A) = 0.3 and P(B) = 0.5, then 𝑃(𝐴 ∩ 𝐵) =? 4) The empirical rule states that, for data having a bell-shaped distribution, the percentage of data values being within one standard deviation of the mean is approximately 5) If a penny is tossed four times and comes up heads all four times, the probability of heads on the fifth trial is ? Answer: 1) Since events are mutually exclusive : P(A n B) =0 and P(A u B) =P(A)+P(B)-P(A n B) =0.3+0.5-0 =0.8 2) since sum of all events is 1 P(E4) =1-0.2-0.3-0.4 =0.1 3) here sign is missing , instead ? is given in question: if it is P(A n B) ; then it is 0 4) percentage of data values being within one standard deviation of the mean is approximately =68 5) probability of heads on the fifth trial is =0.5 Question #5 An important factor in solid missile fuel is the particle size distribution. Significant problems occur if the particle sizes are too large. From production data in the past, it has been determined that the particle size (in micrometers) distribution is characterized by 𝑓(𝑥) = 3𝑥 −4 , for 𝑥 > 1, 𝑓(𝑥) = 0, elsewhere. a) Verify that this is a valid density function. b) Evalute F(x). c) What is the probability that a random particle from the manufactured fuel exceeds 4 micrometers? Answer: Step 1 ∞ a) In order to verify that f is a valid density function, we need to check if the condition ∫−∞ 𝑓(𝑥)𝑑𝑥 = 1 is satisfied. ∞ ∞ ∫−∞ 𝑓(𝑥)𝑑𝑥 = ∫1 3𝑥 −4 𝑑𝑥 = (− 1 𝑥3 ) ∞ 1 = 0 − (− 1 ) (1) 13 =1 In (1) we used the fact: 1 (− 3) = 0 𝑥 ∞ which is true because of: 1 1 (− 3 ) = lim − 3 = 0. 𝑥⇒∞ 𝑥 ∞ 𝑥 The last equality is obtained from the fact: lim 𝑥 3 = ∞ 𝑥⇒∞ Therefore, f truly is a valid density function Step 2 b) Let X be the random variable which represents the particle size in micrometers. By the definition of the cumulative distribution function F, for x>1 we have: 𝐹(𝑥) = 𝑃(𝑋 ≤ 𝑥) 𝑥 = ∫∞ 𝑓(𝑡)𝑑𝑡 𝑥 = ∫1 3𝑡 −4 𝑑𝑡 1 = (− 3) =− 𝑡 1 𝑥3 =1− 𝑥 1 1 − (− 3) 1 1 𝑥3 Step 3 c) We need to find the probability P(X>4) . 𝑃(𝑋 > 4) = 1 − 𝑃(𝑋 ≤ 4) =1-F(4) = 1 − (1 − =1−1+ = 1 1 43 1 ) 64 64 Question #6 Suppose f(x) = 0.125x for 0 < x < 4 . determine the mean and variance of X. Answer: Mean [E(x)] of x The formula for mean is given by, 𝐸(𝑥) = ∫𝑥 𝑓(𝑥)𝑑𝑥 0 = ∫4 𝑥 ∗ 0.125𝑥𝑑𝑥 0 = ∫4 𝑥 2 ∗ 0.125𝑥𝑑𝑥 0 = 0.125 = ∫4 𝑥 2 𝑑𝑥 = 0.125[(𝑥 3 )/(3)]40 {since ∫ 𝑥 𝑛 𝑑𝑥 = (𝑥 (𝑛+1) )/(𝑛 + 1) } = 0.125[(43 )/(3) − 0] = 0.125((64)/3) E(x)=8/3 or 2.667