The Normal Distribution

CHARACTERISTICS OF NORMAL DISTRIBUTION The entire family of normal probability distribution is differentiated by its mean and standard deviation The highest point on the normal curve is at mean which also happens to be median and mode The normal curve is symmetric and the tails of the curve extend to infinity in both directions and do not touch the horizontal axis The Standard deviation decides the width of the curve. Larger the values, wider the curve i.e. indicates more variability

If X is a continuous random variable, then its probability function f(x) also called Probability Density Function (pdf) must satisfy two conditions: a) f(x) 0 and b) The total area under the curve is 1. So, The probability that a variable assumes value between a & b is given by P(a X b) = Area (Density) under f(x) between a & b

NORMAL PROBABILITY DISTRIBUTIONNormal Probability Density function is given by:f(x) = (1/2) e-(x-)2/22Where = mean, = std. deviation = 3.1416e = 2.71828

STANDARD NORMAL PROBABILITY DISTRIBUTIONf(z) = (1/2) e-z2/2 = 0, = 1 for a standard normal probability distributionThe table is generated from this function and is generally called as z-table (standard normal probability distribution table).

STANDARD NORMAL PROBABILITY DISTRIBUTION - GENERALIZED = 100 = 30 = 60 = 351001306095

NORMAL PROBABILITY DISTRIBUTIONThe average amount parents spent on childrens uniform of a particular school was Rs.600 with a standard deviation of Rs.100. a) What is the probability that amount spent on randomly selected child is more than Rs.600?b) What is the probability that amount spent on randomly selected child is between Rs.600 & Rs.750?

NORMAL PROBABILITY DISTRIBUTIONc) What is the probability that a randomly chosen student spends more than Rs.800?d) What is the probability that a randomly chosen student spends between Rs.650 and Rs.750?e) What is the probability that a randomly chosen student lesser than Rs.670?f) What is the probability that a randomly chosen student spends between 520 and 680?

NORMAL PROBABILITY DISTRIBUTIONg) If a students spend is so high that it falls in the upper 10% of the spend, how much a student must be spending in Rupees?

GREAR TIRE PROBLEMGrear Tire Company has just developed a new steel-belted radial tire and has estimated the mean tire mileage is = 36500 miles and s.d. = 5000 miles. What is the probability that the tire mileage will exceed 40000?. Assume that Grear is considering a guarantee that will provide a discount on replacement tires if the original tires do not exceed the mileage stated in the guarantee. What should the guarantee mileage be if Grear wants no more than 10% of the tires to be eligible for the discount guarantee?.

INTERNET USAGEIn Jan, An average Indian spent an average of 77 hours logged on to the Internet while at work. Assume the times are normally distributed and that the standard deviation is 20 hours. a) What is the probability a randomly selected worker spent fewer than 50 hrs logged on to the internet?b) What % of workers spent more than 100 hrs logged on to the internet?c) A person is classified as heavy user if he/she is in the upper 20% of usage. How many hrs to be spent to be called so?

NORMAL APPROXIMATION OF BINOMIAL PROBABILITIESWhen the number of trials become large, it is difficult to use Binomial distribution.When the number of trials 20, np 5 and nq 5, we can use normal distribution for approximation of binomial distributionWhen using normal distribution to approximate binomial distribution, we use = np and = npq

NORMAL APPROXIMATION OF BINOMIAL PROBABILITIESABC co. has a history of making errors in 10% of its invoices. A sample of 100 invoices has been taken, and we want to know the probability that 12 invoices will contain errors. Continuity correction factor: Add and subtract 0.5 from the value.