Question

Suppose that the scores on a statewide standardized test are normally distributed with a mean of 70 and a standard deviation of 4. Estimate the percentage of scores that were(a) between 66 and 74. %(b) above 82. %(c) below 66. %(d) between 62 and 82. %

Answer 1

Remember that

z =(x - μ)/σ

so

we have

μ=70

σ=4

Part a

For x=66 ------> z=(66-70)/4 ------> z=-1

For x=74 -----> z=(74-70)/4 -----> z=1

using a z table scores

p=0.68269

convert to percentage

p=68%

Part b

above 82

For x=82 ------> z=(82-70)/4 -----> z=3

using a z table scores

p=0.001350

convert to percentage

p=0.14%

Part c

below 66

z=(66-70)/4=-1

using a z table scores

p=0.15866

p=15.87%

Part d

between 62 and 82

For x=62 -----> z=(62-70)/4 ----> z=-3

For x=82 -----> z=(82-70)/4 ----> z=3

p=0.9973

p=99.73%