More on Normal Distribution

Thursday, 15 August 2019

7:03 PM

Machine generated alternative text: Lecture 4 Normal distribution: examples Example 1.16 p.38 (textbook) The time it takes a driver to react to the brake light on a decelerating vehicle follows a Normal distribution having parameters = 1.25 sec and a = 0.46 sec. In the long run, what proportion of reaction times will be between 1 and 1.75 sec? (Hint: = 0.8621 , = 0.2946.) We have X N (1.25, 0.46) and we desire P(l X 1.75) . We have that f X S 1.75) = 1.75) - 1) The probabilities can be obtained from Matlab. .25 JV(O, 1). Then Alternatively, we know that Z 0.46 x- 1.25 1.75- 1.25 1.75) = P — IP(Z 1.09) — 0.8621 Similarly, 1) = -0.54) = = 0.2946, so that f X S 1.75) = 0.8621 - 0.2946 = 0.5675 MATH2099/MATH2859 (Statistics) Dr Jia Deng Solutions - Lecture 4 8/10

Machine generated alternative text: 5. Special random variables The Normal distribution 5.6 Normal random variables A random variable is said to be normally distributed with parameters and a (a > 0), i.e. if its probability density function is given by f(x) — 2 1 (x—g) e 20 2T(J Unfortunately, no closed form exists for 2 x 1 202 2T(J —00 Important remark: Be careful! Many sources use the alternative notation X N ( p, 02) + in the textbook and in Matlab, the notation N (g, a) is used, so we adopt it in these slides as well MATH2099/2859 (Statistics) Dr Jia Deng Term 2, 2019 - Lecture 4 42/72

 

Machine generated alternative text: 5. Special random variables 5.6 Normal random variables The Standard Normal distribution The Standard Normal distribution is the Normal distribution with = 0 and = 1. This yields f(x) — 1 27T Usually, in this situation, the specific notation f(x) qb(x) and is used, and a standard normal random variable is usually denoted Z. It directly follows from the preceding that var(Z) = 1 (= sd(Z)) and MATH2099/2859 (Statistics) Dr Jia Deng Term 2, 2019 - Lecture 4 45/72

 

 

Machine generated alternative text: Lecture 4 Normal distribution: examples Example The actual amount of instant coffee that a filling machine puts into "4-ounce" jars may be looked upon as a random variable having a normal distribution with a = 0.04 ounce. If only 2% of the jars are to contain less than 4 ounces, what should be the mean fill of these jars? (Hint: = 0.02.) Let X denote the actual amount of coffee put into the jar by the machine We have X N (p, 0.04), with such that P(X 4) — — 0.02 Solutions - Lecture 4 Hence, 0.02 = 4) = P 0.04 0.04 According to the hint, P(Z —2.05) = 0.02 0.04 We conclude that 0.04 ¯ MATH2099/MATH2859 (Statistics) —2.05, that is, = 4 + 0.04 x 2.05 = 4.082 ounces Dr Jia Deng 9/10

 

Machine generated alternative text: Lecture 4 Some further properties of the Normal distribution Example (ctd.) If = 40 min, = 50 min and = 60 min, and = 10 min2, = 12 min2 and = 14 min2, what is the probability that the full task would take less than 160 min? (Hint: = 0.9525.) From the above, we have -6) Solutions - Lecture 4 X (150, 36 Hence, 160) MATH2099/MATH2859 (Statistics) 160 - 150 6 = 1.67) = 0(1.67) = 0.9525 Dr Jia Deng 10/10

 

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