(i) To find the proportion of students who scored between 60 and 85, we first need to convert these scores to z-scores. The z-score is calculated as (X - μ) / σ, where X is the score, μ is the mean, and σ is the standard deviation.

For X = 60, the z-score is (60 - 70) / 15 = -0.67.
For X = 85, the z-score is (85 - 70) / 15 = 1.

Looking up these z-scores in a standard normal distribution table, we find that the proportion of students who scored less than 60 is 0.2514 and the proportion who scored less than 85 is 0.8413.

Therefore, the proportion of students who scored between 60 and 85 is 0.8413 - 0.2514 = 0.5899 or 58.99%.

(ii) The question seems to have a typo. A score of 9 is extremely low given the mean of 70 and standard deviation of 15. If it's not a typo, the z-score for X = 9 is (9 - 70) / 15 = -4.07. This is off the charts of the standard normal distribution table, so we can say that virtually no students scored above 9.

(iii) For X = 50, the z-score is (50 - 70) / 15 = -1.33. Looking this up in the standard normal distribution table, we find that the proportion of students who scored less than 50 is 0.0918 or 9.18%.

Question

(i) To find the proportion of students who scored between 60 and 85, we first need to convert these scores to z-scores. The z-score is calculated as (X - μ) / σ, where X is the score, μ is the mean, and σ is the standard deviation.

For X = 60, the z-score is (60 - 70) / 15 = -0.67. 
For X = 85, the z-score is (85 - 70) / 15 = 1.

Looking up these z-scores in a standard normal distribution table, we find that the proportion of students who scored less than 60 is 0.2514 and the proportion who scored less than 85 is 0.8413.

Therefore, the proportion of students who scored between 60 and 85 is 0.8413 - 0.2514 = 0.5899 or 58.99%.

(ii) The question seems to have a typo. A score of 9 is extremely low given the mean of 70 and standard deviation of 15. If it's not a typo, the z-score for X = 9 is (9 - 70) / 15 = -4.07. This is off the charts of the standard normal distribution table, so we can say that virtually no students scored above 9.

(iii) For X = 50, the z-score is (50 - 70) / 15 = -1.33. Looking this up in the standard normal distribution table, we find that the proportion of students who scored less than 50 is 0.0918 or 9.18%.

Knowee AI · Accepted Answer

(i) To find the proportion of students who scored between 60 and 85, we first need to convert these scores to z-scores. The z-score is calculated as (X - μ) / σ, where X is the score, μ is the mean, and σ is the standard deviation.

For X = 60, the z-score is (60 - 70) / 15 = -0.67. 
For X = 85, the z-score is (85 - 70) / 15 = 1.

Looking up these z-scores in a standard normal distribution table, we find that the proportion of students who scored less than 60 is 0.2514 and the proportion who scored less than 85 is 0.8413.

Therefore, the proportion of students who scored between 60 and 85 is 0.8413 - 0.2514 = 0.5899 or 58.99%.

(ii) The question seems to have a typo. A score of 9 is extremely low given the mean of 70 and standard deviation of 15. If it's not a typo, the z-score for X = 9 is (9 - 70) / 15 = -4.07. This is off the charts of the standard normal distribution table, so we can say that virtually no students scored above 9.

(iii) For X = 50, the z-score is (50 - 70) / 15 = -1.33. Looking this up in the standard normal distribution table, we find that the proportion of students who scored less than 50 is 0.0918 or 9.18%.

3) Suppose the test scores on the APU entrance exam is normally distributed with a meanof 70 and a standard deviation of 15.(i) What proportion of students scored between 60 and 85.(ii) What proportion of students scored above 9(iii) What proportion of students scored less than 50

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