Question

An education researcher has been studying test anxiety using a particular​ measure, which she administers to students prior to midterm exams. On this​ measure, she has found that the distribution follows a normal curve. The measure has a mean of 13

and a standard deviation of 4.

Using the accompanying normal curve​ table, what percentage of students have scores​ (a) above 22​,

​(b) above

20​,

​(c) above

19​,

​(d) below

19​,

and​

(e) below

4​?

Answer 1

Solution :

Given that ,

mean = $\mu$ = 13

standard deviation = $\sigma$ = 4

(a)

P(x > 22) = 1 - P(x < 22)

= 1 - P[(x - $\mu$ ) / $\sigma$ < (22 - 13) / 4]

= 1 - P(z < 2.25)

= 1 - 0.9878

= 0.0122

percentage = 1.22%

(b)

P(x > 20) = 1 - P(x < 20)

= 1 - P[(x - $\mu$ ) / $\sigma$ < (20 - 13) / 4]

= 1 - P(z < 1.75)

= 1 - 0.9599

= 0.0401

percentage = 4.01%

(c)

P(x > 19) = 1 - P(x < 19)

= 1 - P[(x - $\mu$ ) / $\sigma$ < (19 - 13) / 4]

= 1 - P(z < 1.5)

= 1 - 0.9332

= 0.0668

percentage = 6.68%

(d)

P(x < 19) = P[(x - $\mu$ ) / $\sigma$ < (19 - 13) / 4]

= P(z < 1.5)

= 0.9332

Percentage = 93.32%

(e)

P(x < 4) = P[(x - $\mu$ ) / $\sigma$ < (4 - 13) / 4]

= P(z < -2.25)

= 0.0122

Percentage = 1.22%