Question

Based on data from a​ college, scores on a certain test are normally distributed with a mean of 1532 1532 and a standard deviation of 316 316. LOADING... Click the icon to view the table with standard scores and percentiles for a normal distribution. a. Find the percentage of scores greater than 1848 1848. nothing ​% ​ (Round to two decimal places as​ needed.) b. Find the percentage of scores less than 900 900. nothing ​% ​ (Round to two decimal places as​ needed.) c. Find the percentage of scores between 1374 1374 and 2006 2006. nothing ​% ​ (Round to two decimal places as​ needed.)

Standard score   Percent
-3.0   0.13
-2.5   0.62
-2   2.28
-1.5   6.68
-1   15.87
-0.9   18.41
-0.5   30.85
-0.1   46.02
0   50.00
0.10   53.98
0.5   69.15
0.9   81.59
1   84.13
1.5   93.32
2   97.72
2.5   99.38
3   99.87
3.5   99.98

Answer 1

Given that

mean = 1532

standard deviation = 316

(A) we can write 1848 as 1532+(1*316), i.e. mean + 1*Sd

so, it is one standard deviation above the mean

using the given table, 84.13% of scores are below 1 standard deviation

So, % of scores above 1848 = 100 - 84.13 = 15.87%

(B)

we can write 900 as 1532-(2*316), i.e. mean - 2*Sd

so, it is 2 standard deviation below the mean

using the given table, 2.28% of scores are below 2 standard deviation

So, % of scores below 900 = 2.28%

(C)

we can write 2006 as 1532+(1.5*316), i.e. mean + 1.5*Sd

so, it is 1.5 standard deviation above the mean

using the given table, 93.32% of scores are below 1.5 standard deviation

and

we can write 1374 as 1532-(0.5*316), i.e. mean - 0.5*Sd

so, it is 0.5 standard deviation below the mean

using the given table, 30.85% of scores are below 0.5 standard deviation

So, % of scores between 1374 and 2006 = 93.32 - 30.85 = 62.47%