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Answer :
We are given the average grade as 85 and the standard deviation as 15 points. Using these, we need to find out various percentages of students in the class based on the given conditions, which are explained below:The mean of the class is 85, and standard deviation is 15.
The score which is less than 65 will be calculated using the z-score formula as:
z = (x - μ) / σ
Where x = 65, μ = 85, and σ = 15Substituting the values, we have
z = (65 - 85) / 15z = -1.33
The probability of the score being less than 65 is given by the probability of getting a z-score less than -1.33. Using the z-table, we can find the area as 0.0912, which can be multiplied by the total number of students to get the number of students that got a failing grade.Approximately 4 students received a failing grade (less than 65). We are given the results of the first test in a statistics class. We have to find out various percentage values based on the data given in the question. The mean value is 85, and the standard deviation is 15 points. By using the formula for z-score, we can find out the percentage of students who got grades less than or greater than a certain value. For instance, to find out the percentage of students who scored between 70 and 91, we first need to calculate the z-score for these values.The z-score for a value of 70 is:
z = (x - μ) / σ= (70 - 85) / 15= -1
The z-score for a value of 91 is:
z = (x - μ) / σ= (91 - 85) / 15= 0.4
We then find the probability of getting a value between these two z-scores. We use the standard normal distribution table to find this value. We know that the probability of getting a z-score between -1 and 0.4 is 0.4222. This value multiplied by the total number of students will give us the number of students who scored between 70 and 91. We can use a similar method to find out the number of students that received a score whose z-score was -.70 or less.To find the grade required to be in the top 30%, we first need to find out the z-score that corresponds to this percentile. We know that the area to the left of a z-score of 0.52 is 0.6997. Therefore, the area to the right of this z-score is 0.3003, which corresponds to the top 30% of the class. We then use the formula for z-score to find the corresponding grade value as:
z = (x - μ) / σ0.52 = (x - 85) / 15x = (0.52 * 15) + 85x = 93.8
Therefore, the grade required to be in the top 30% is 93.8.To find the grade required to be in the top 22%, we first need to find out the z-score that corresponds to this percentile. We know that the area to the left of a z-score of 0.81 is 0.7902. Therefore, the area to the right of this z-score is 0.2098, which corresponds to the top 22% of the class. We then use the formula for z-score to find the corresponding grade value as:
z = (x - μ) / σ0.81 = (x - 85) / 15x = (0.81 * 15) + 85x = 96.15
Therefore, the grade required to be in the top 22% is 96.15.
To summarize, we used the given mean and standard deviation values to find out various percentages of students based on different conditions. We calculated the number of students that received a failing grade, the number of students that received a grade between 70 and 91, the number of students that received a score whose z-score was -.70 or less, the grade required to be in the top 30%, and the grade required to be in the top 22%.
To learn more about z-table visit:
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