Question
Download Solution PDFA test was given to 360 students of Class X and a normal distribution of scores was obtained. The mean of scores was 45 with a standard deviation of 9 points. The percentage of students who scored outside the limits of scores 36 - 54 would be
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFKey Points
Normal Probability Curve: It is a bell-shaped curve having highest point at mean which is symmetrical along the vertical line drawn at the mean. It is used to interpret the percentile or the percentage of no of cases for respective value of z scores.
Mean: Mean (average) is sum of all quantities divided by no of quantities.
Standard deviation: It is used to measure how deviated or dispersed the numbers are with respect to the mean in the same set of data.
Percentile Rank: It is a measure to find how many values are below the score. It is basically the rank of the score.
Z score: Z score is a value calculated through the above formula, used to interpret percentile of a given score or value.
Important Points
Given: n = 360, Mean Score (x̄) = 45 , σ = 9
To find: percentage of students who scored outside the limits of scores 36 – 54
Formula: Z = (x−x̄)/σ
Calculation:
To find out the percentage of students in a limit, we need to calculate z score
At x=36, Z = (36−45)/9 = -1
At x= 54, Z = (54−45)/9 = 1
From Normal Probability curve, when z score is -1 to 1, percentage/ percentile rank is =68.26%
Percentage of students who scored outside the limits of scores 36 – 54 = Total - 68.26%= 100% - 68.26% = 31.74% = 32% approx
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