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UIUC STAT 400 - 400Ex3_3ans

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STAT 400 Lecture AL1 Examples for 3.3 Spring 2015 Dalpiaz 1. At Initech , the salaries of the employees are normally distributed with mean µ = $36,000 and standard deviation σ = $5,000. a) Mr. Smith is paid $42,000. What proportion of the employees of Initech are paid less that Mr. Smith? P( X < 42,000 ) = −<000,5000,36000,42ZP = P( Z < 1.20 ) = 0.8849. b) What proportion of the employees have their salaries over $40,000? P( X > 40,000 ) = −>000,5000,36000,40ZP = P( Z > 0.80 ) = 1 – 0.7881 = 0.2119. c) Suppose 10 Initech employees are randomly and independently selected. What is the probability that 3 of them have their salaries over $40,000? Let Y = number of employees (out of 10) who have salaries over $40,000. Then Y has Binomial distribution, n = 10, p = 0.2119 ( see (b) ). P( Y = 3 ) = ( ) ( )737881.02119.0310C ⋅⋅ = 0.2156. d) What proportion of the employees have their salaries between $30,000 and $40,000? P( 30,000 < X < 40,000 ) = −<<−000,5000,36000,40Z000,5000,36000,30P = P( – 1.2 < Z < 0.80 ) = 0.7881 – 0.1151 = 0.6730.e) Mrs. Jones claims that her salary is high enough to just put her among the highest paid 15% of all employees working at Initech . Find her salary. Need x = ? such that P( X > x ) = 0.15. ( area to the right is 0.15 ) First, need z = ? such that P( Z > z ) = 0.15. z = 1.04. X = µ + σ Z. x = 36,000 + 5,000 × 1.04 = $41,200. f) Ms. Green claims that her salary is so low that 90% of the employees make more than she does. Find her salary. Need x = ? such that P( X > x ) = 0.90. ( area to the right is 0.90 ) First, need z = ? such that P( Z > z ) = 0.90. z = – 1.28. X = µ + σ Z. x = 36,000 + 5,000 × ( – 1.28 ) = $29,600. 2. Suppose that the lifetime of Outlast batteries is normally distributed with mean µ = 240 hours and unknown standard deviation. Suppose also that 20% of the batteries last less than 219 hours. Find the standard deviation of the distribution of the lifetimes. Need σ = ? Know P( X < 219 ) = 0.20. First, need z = ? such that P( Z < z ) = 0.20. z = – 0.84. X = µ + σ Z. 219 = 240 + σ × ( – 0.84 ). – 21 = σ × ( – 0.84 ). σ = 25 hours.Let X be normally distributed with mean µ and standard deviation σ. Then the moment-generating function of X is M X ( t ) = 2 σ μ 22 tte+. M X ( t ) = E ( e t X ) = ( )∫∞∞−−−⋅ dxxxtee2 2 σμ 212 σπ = ()∫∞∞−−+⋅ dzzztee2 2 21σμπ = ()∫∞∞−−−+⋅ dztzttee2 2 2 22 σ 21σμπ = 2 22 σμ tte+, since ()2 2 σ21tze−−π is the probability density function of a N ( σ t , 1 ) random variable. Let Y = a X + b. Then M Y ( t ) = e b t M X ( a t ). Therefore, Y is normally distributed with mean a µ + b and variance a 2 σ 2 ( standard deviation | a | σ ). 1. (continued) g) All Initech employees receive a memo instructing them to put away 4% of their salaries plus $100 per month ( $1,200 per year ) in a special savings account to supplement Social Security. What proportion of the employees would put away more than $3,000 per year? Y = 0.04 X + 1,200. P( Y > 3,000 ) = ? Y > 3,000 ⇔ X > 45,000. P( X > 45,000 ) = −>000,5000,36000,45ZP = P( Z > 1.80 ) = 1 – 0.9641 = 0.0359.OR µ Y = 0.04 × 36,000 + 1,200 = $2,640, σ Y = 0.04 × 5,000 = $200. P( Y > 3,000 ) = −>200640,2000,3ZP = P( Z > 1.80 ) = 1 – 0.9641 = 0.0359. 3. Suppose the average daily temperature [in degrees Fahrenheit] in June in Anytown is a random variable T with mean µ T = 85 and standard deviation σ T = 7. The daily air conditioning cost Q , in dollars, for Anytown State University, is related to T by Q = 120 T + 750. Suppose that T is a normal random variable. Compute the probability that the daily air conditioning cost on a typical June day for the university will exceed $12,210. Q has Normal distribution. µ Q = 120 µ T + 750 = 120 ⋅ 85 + 750 = $10,950. () ( )222 22 7 120 120TQσσ ⋅⋅ == = 840 2. σ Q = $840. P( Q > 12,210 ) = −>8409501021012ZP,, = P( Z > 1.50 ) = 1 – Φ ( 1.50 ) = 1 – 0.9332 = 0.0668. OR 12,210 = 120 T + 750. ⇔ T = 95.5. P( Q > 12,210 ) = P( T > 95.5 ) = −>785595ZP. = P( Z > 1.50 ) = 1 – Φ ( 1.50 ) = 1 – 0.9332 = 0.0668.4. Consider a random variable X with the moment-generating function M X ( t ) = e 3 t + 8 t 2 = exp ( 3 t + 8 t 2 ). a) Find P ( X > 0 ). Normal distribution: M X ( t ) = 2 22 σμ tte+. ⇒ X has a Normal distribution with µ = 3 and σ 2/ 2 = 8. ⇒ E ( X ) = µ = 3, Var ( X ) = σ 2 = 16. σ = 4. P ( X > 0 ) = −>430ZP = P ( Z > – 0.75 ) = 1 – Φ ( – 0.75 ) = 1 – 0.2266 = 0.7734. a) Find P ( – 1 < X < 9 ). P ( – 1 < X < 9 ) = −<<−−439Z431P = P ( – 1.00 < Z < 1.50 ) = Φ ( 1.50 ) – Φ ( – 1.00 ) = 0.9332 – 0.1587 = 0.7745. ___________________________________________________________________________ EXCEL: = NORMSDIST( z ) gives P( Z ≤ z ) = Φ( z ) = NORMSINV( p ) gives z such that P( Z ≤ z ) = p = NORMDIST( x , µ , σ , 1 ) gives P( X ≤ x ), where X is N( µ , σ 2 ) = NORMDIST( x , µ , σ , 0 ) gives f ( x ), p.d.f. of N( µ , σ 2 ) = NORMINV( p , µ , σ ) gives x such that P( X ≤ x ) = p, where X is N( µ , σ 2 )5.* Show that the odd moments of N ( 0, σ 2 ) are zero and …


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UIUC STAT 400 - 400Ex3_3ans

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