Students of a class were given an aptitude test. Their marks were found to be normally distributed with mean 60 and standard deviation 5. What percentage of students scored. i) More than 60 marks (ii) Less than 56 marks (iii) Between 45 and 65 marks Solution: Given that mean = m = 60 and standard deviation = s = 5 i) The standard normal varaiate Z = s x m- If X = 60, Z = s x m- = 0 5 6060 = - \P(x > 60) = P(z > 0) = P(0 < z < ¥ ) = 0.5000 Hence the percentage of students scored more than 60 marks is 0.5000(100) = 50 % 0.5 - ¥ z = 0 z > 0 + ¥ 96 ii) If X = 56, Z = 8.0 5 4 5 6056 -= - = - P(x < 56) = P(z < -0.8) = P(- ¥ < z < 0) – P(-0.8 < z < 0) (by symmetry) = P(0 < 2 < ¥) – P(0 < z < 0.8) = 0.5 - 0.2881 (from the table) = 0.2119 Hence the percentage of students score less than 56 marks is 0.2119(100) = 21.19 % iii) If X = 45, then z = 3 5 15 5 6045 -= - = - X = 65 then z = 1 5 5 5 6065 == - P(45 < x < 65) = P(-3 < z < 1) = P(-3 < z < 0 ) + P ( 0 < z < 1) 0.2119 - ¥ z= -0.8 z=0 + ¥ 0.83995 - ¥ z= -3 z=0 z=1 + ¥ 97 = P(0 < z < 3) + P(0 < z < 1) ( by symmetry) = 0.4986 + 0.3413 (from the table) = 0.8399 Hence the percentage of students scored between 45 and 65 marks is 0.8399(100) = 83.99 % Example 22: X is normal distribution with mean 2 and standard deviation 3. Find the value of the variable x such that the probability of the interval from mean to that value is 0.4115 Solution: Given m = 2, s = 3 Suppose z1 is required standard value, Thus P (0 < z < z1) = 0.4115 From the table the value corresponding to the area 0.4115 is 1.35 that is z1 = 1.35 Here z1 = s x m- 1.35 = 3 - 2x x = 3(1.35) + 2 = 4.05 + 2 = 6.05