The net promoter score measures the quality of product and level of customer satifaction (0 lowest, 10 highest). A particular industry has an average score of μ=7 with standard deviation of σ= 4. What is Q3 (75th percentile) of average scores from a sample of n= 30 businesses within this industry?
Solution :
Given that,
mean = = 7
standard deviation = = 4
n = 30
= 7
= / n = 4 / 30
P(Z < z) = 0.75
P(Z < 0.67) = 0.75
z = 0.67
= z * + = 0.67 * 4 / 30 + 7 = 7.5
75th percentile = 7.5
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