the mean intelligence level of a group of students is 90 with standard deviation 20. Assuming that the intelligence level is normally distributed, find the percentage of students with intelligence over 100 {given, area between Z=0 to Z=0.5 is 0.1915]
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Answer:
100 is the average, so by symmetry, exactly 50% of the population has an IQ score of 100 or better. 115 is one standard deviation above the mean, i.e., z = 1.0.
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Solution.
z-value Probability (area)
2.00 0.4772
2.50 0.4938
3.00 0.4987 (almost 50%)
Step-by-step explanation:
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