Question

Use the given z-score/percentile table to answer the following question: If scores on a test are normally distributed with mean 48 and standard deviation 10, what percentage are above 62? z-score 1.1 1.2 1.3 1.4 1.5

percentile 86.43 88.49 90.32 91.92 93.32

Answer 1

Solution :

Given that,

mean = = 48

standard deviation = =10

P(x > 62) = 1 - P(x< 62)

= 1 - P[(x -) / < (62-48) /10 ]

= 1 - P(z <1.4 )

Using z table

= z score 1.4 percentage 91.92