Question

EXAMPLE I: Consider the weights of 18 month old boys in the U.S. According to published growth charts, the average weight is approximately 11.8 kg with standard deviation of 1.28 kg. Calculate the percentage of 18 month old boys in the U.S. who weigh between 10.5 kg and 14.4 kg. Solution: So X- N(11.8, 1.28). We need P(10.5<X<14.4) Here: Lower boundary- 10.5 Second: Upper boundary 14.4 Third: u-11.8 Fourth: 6-1.28 Now normaledf(10.5, 14.4, 11.8, 1.28) 824 We want to convert this number to a percentage, so our answer is 82.4 % EXAMPLE 2: Adult 1Qs are normally distributed with u-100 and a-15. a) Find the probability that a randomly selected IQ is less than 112. b) Find the probability that a randomly selected IQ is at least 122. That is, find P(x 122). Solution: Let X be Adult IQs and so X - N(100, 152). We like to find a) P(x<112). Since this is a continuous function, we have P(x < 112)-P(xS 112) PROBLEM: no lower bound!!!! Note: The -9999 is used as the Ib to be at least 5 standard deviations from the mean. b) P(X> 122)- normaledf(122,9999,100,15)-.0712. normaledf(-9999,112,100,15)-.7881. EXAMPLE 3: If a random variable has the normal distribution with u-82.0 and o-4.8, find the probabilities that it will take on a value (a) less than 89 fCX9)har n COF (99, 5,31 (b) greater than 78 (c) between 83 and 88 (d) between 73 and 90 EXAMPLE 4: Let X N(-5, 9). Find PCX<2) 955912 P(XI> 1) 7816 -,9) (1,9999-,9) 2624 EXAMPLE 5: We have a national examination given to a large group of students and we know the data is normally distributed. The average for the data is 300 with a standard deviation of 50. What is the percentile score of someone who earns 215 on the exam? Solution: Identify the average (300), the standard deviation (50), and the raw score (215). EXAMPLE 6: 1Qs are normally distributed with a mean of 100 and a standard deviation of 16. Determine the percentage of people who have IQs between 115 and 140. 16-1610 16.2% nn(lo0) EXAMPLE 7: As reported the times of the finishers in the New York City 10-km run are normally distributed with mean 61 minutes and standard deviation 9 minutes. (a) Determine the percentage of finishers with times between 50 and 70 minutes. (b) Determine the percentage of finishers with times less than 75 minutes.

Answer 1

(Example 1 and Example 2 are already solved, so I am solving the rest.)

(3) X ~ N(82, 4.8).
(a) normalcdf(0,89,82,4.8) = 0.9276
(b) normalcdf(78,9999,82,4.8) = 0.7977
(c) normalcdf(83,88,82,4.8) = 0.3118
(d) normalcdf(73,90,82,4.8) = 0.9218

(4) X ~ N(-5,9)
(a) normalcdf(-2,2,-5,9) = 0.1511
(b) 1 - normalcdf(-1,1,-5,9) = 0.9241

(5) X ~ N(300,50)
normalcdf(0,215,300,50) = 0.0446 = 4.46th percentile = 4th percentile (approx.)

(6) X ~ N(100,16)
normalcdf(115,140,100,16) = 0.1680 = 16.8%

(7) X ~ N(61,9)
(a) normalcdf(50,70,61,9) = 0.7305 = 73.05%.
(b) normalcdf(0,75,61,9) = 0.9401 = 94.01%.