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The quiz result for our class with a normal distribution has the below features:Mean 73Standard Deviation 8Pls find:- (i) probability of getting 91 OR LESS by the students;(ii) percentage of students getting BETWEEN 65 AND 89;(iii) percentage of students getting BETWEEN 81 AND 89.(iv) probability of getting... 顯示更多 The quiz result for our class with a normal distribution has the below features: Mean 73 Standard Deviation 8 Pls find:- (i) probability of getting 91 OR LESS by the students; (ii) percentage of students getting BETWEEN 65 AND 89; (iii) percentage of students getting BETWEEN 81 AND 89. (iv) probability of getting NO MORE THAN 50. (v) probability AND percentage of getting MORE THAN 90. p.s. Pls kindly explain with enlarged wording (因為大近視和老花). Tks so much!

最佳解答:

i) 91 is 2 standard deviations above the mean and hence the required prob. is: P(z <= 2) = 0.9772 ii) 65 is 1 standard deviation below the mean and 89 is 1.75 standard deviations above the mean and hence the required prob. is: P(-1 <= z <= 1.75) = 0.8012, with percentage = 80.12 iii) 81 is 1 standard deviation above the mean and 89 is 1.75 standard deviations above the mean and hence the required prob. is: P(1 <= z <= 1.75) = 0.1186, with percentage = 11.86 iv) 50 is 2.875 standard deviations below the mean and hence the required prob. is: P(z <= -2.875) = 0.00205 v) 90 is 1.875 standard deviations above the mean and hence the required prob. is: P(z > 1.875) = 0.0304

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