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Normal Distribution Normal Distribution: Suppose π is the set of continuous set of data with mean π and variance π 2 and the graph of p.d.f of this set of data is bell shape then X is said to be normally distributed. π and π 2 are said to be parameter of π. If X is normally distributed then it can be written as π~(π, π 2 ) Examples. π~π(3, 0.9) π~π(14, 4) π~π(78, 25) Curve of normal distribution: The normal distribution curve has the following properties. ο· It is bell shaped. ο· It is symmetrical about mean. ο· It extends from β β π‘π + β. ο· The total area under the curve is 1. ο· Approximately 99.8% of the distribution lies within 3 standard deviation of mean. Standard normal variable: If a normal has mean 0 and variance 1 then it is called standardized normal distribution and it is denoted by Z. π~(0, 1) Probability density function of standard normal distribution: The p.d.f of standard normal distribution is denoted by β (π§) and it gives the area under the curve before the value of π§. Use of standard normal table: ο· To find the value of β (π§) at given value of z the number must be in three decimal places. ο· First decimal place is looked in first column and second decimal place is looked in first row. Find the crossed value. ο· Third decimal place is looked in βColumn of ADDβ. This value is added into last two digits of crossed value. Q.1 Find the value of function β (π§), i. β (1.234) iii. β (1.6789) ii. β (2.5) iv. β (2.45) Relation between area under the graph and probability of standardized normal Variable: Basic Rules: ο· P(Z < π) = β (π§) ο· P(Z > π) = P(Z < βa) ο· β (βπ) = π β β (π) ο· P(a < Z < b) = β (π) β β (π) ο· P(|Z| < π) = P(βa < Z < a) ο· P(|Z| > π) = P(Z < βπ ππ Z > π) = π(π < βπ) + π(π > π) Q.2 If π~(0, 1), find i. P(Z < 0.16) ii. P(Z < β1.346) iii. P(Z > 2.9) iv. P(Z < 0.1362) v. P(|Z| < 1.12) vi. P(|Z| > 1.89) Use of standard normal table to find β βπ (π§): ο· Look the probability in main body of the table. If is not appear then look the value is less than an very close to it. ο· Find in which row and column it is lying. ο· Column and row will make first and second decimal places of given value. ο· The column of βADDβ will make third decimal places. Compiled By : Sir Rashid Qureshi www.levels.org.pk Note: To find the value of variable from given probability first decide the sign of value of variable by using the following information. If π(π < π) = < 0.5 ππ π(π > π) = > 0.5 then π is negative , in all other cases the value of π is positive. Q.3 Find the value of π such that i. P(Z < a) = 0.5685 ii. P(Z < a) = 0.4321 iii. P(Z > π) = 0.7654 iv. P(Z < π) = 0.9833 v. P(|Z| < π) = 0.4536 vi. P(|Z| > π) = 0.112 Use of standardized normal table for any normal distribution: A normal distribution π~(π, π 2 ) can be changed into standardized normal distribution π~(0, 1) by the following formula. πβπ π= π Q.4 The r.v π is such that π~(50, 8) then find i. P(48 < π < 54) ii. P(52 < π < 55) iii. P(46 < π < 49) iv. P(|X β 50| < β8) Q.5 Packages from a packing machine have a mass which is normally distributed with mean 200g and standard deviation 2g. Find the probability that a package from the machine weighs i. Less than 197g, ii. More than 200.5g, iii. Between 198.5g and 199.5g, Q.6 A certain type of the cabbage has a mass which is normally distributed with mean 1 and standard deviation 0.15 kg. Ina lorry load of 800 of these cabbages, estimate how many will have mass i. Greater than 0.79kg, ii. Less than 1.13kg, iii. Between 0.85kg and 1.15 kg, iv. Between 0.75kg and 1.29kg. De-standardizing: Sometimes the probability of a normal distribution is given and the value of variable has to find then de-standardization is used. i. To de-standardizing first change the given normal distribution into standardized normal distribution. ii. By analyzing the value of probability find the sign of unknown variable.( it will be more easy if this quantity is supposed new variable. Q.7 If π~π(24, 9) and P(X > a) = 0.974, find the value of π. Q.8 The distance in meters that a ball can be thrown by pupils at a particular school follows a normal distribution with mean 35.0 and standard deviation 11.6 m. i. Find the probability that a randomly chosen pupil can throw a ball between 30 m and 40 m. ii. The school gives a certificate to the 10 % of pupils who throw further than a certain distance. Find the least distance that must be thrown to qualify the certificate. Compiled By : Sir Rashid Qureshi www.levels.org.pk Approximation of binomial to normal distribution: under certain circumstances the normal distribution can be used as an approximation to the binomial distribution. A Binomial distribution can be approximated if it satisfies the conditions. π~π΅(π, π) Then the it can be approximated ifππ > 5 πππ ππ > 5. Examples. Determine which of the following Binomialsβ can be approximated normal distribution or not. ο· π~π΅(20, 0.3) ο· π~π΅(26, 0.5) ο· π~π΅(14, 0.4) Method to approximation i. Find the mean and variance of the binomial distribution. If π~π΅(π, π) then πΈ(π) = ππ and π(π) = πππ. ii. π~π΅(π, π) then π~π(ππ, πππ ) iii. To find the probability use the standardized normal distribution and probability distribution function. iv. Continuity correction. Binomial distribution Normal distribution π(π < 3) π(π < 2.5) π(π β€ 3) π(π β€ 3.5) π(π > 3) π(π > 3.5) π(π β₯ 3) π(π β₯ 2.5) π(π = 3) π(2.5 < π < 3.5) Q.9 Find the probability of obtaining between 4 and 7 heads inclusive with 12 tossess ofa fair coin, i. Using the binomial distribution. ii. Using the normal approximation to the binomial distribution. Q.10 The length of certain items follows a normal distribution with mean π and standard deviation 6 cm. It is known that 4.78% of the items have a length greater than 82 cm. find the value of mean π. Q.11 The masses of articles produced in a particular workshop are normally distributed with mean π and standard deviation π. 5% of the articles have a mass greater than 85g and 10% have a mass less than 25g. Find the values of π and π. Compiled By : Sir Rashid Qureshi www.levels.org.pk Compiled By : Sir Rashid Qureshi www.levels.org.pk