Answer:
a) P ( X < 91 ) = 0.9927
b) P ( 65 < X < 91 ) = 0.6585
c) P(81 < X < 89 ) =0.0781
d) X = 83.8
Step-by-step explanation:
Given:
- Mean of the distribution u = 69
- standard deviation sigma = 9
Find:
a. what is the probability of getting a grade of 91 or less on this exam?
b. What percentage of students scored between 65 and 89?
c. What percentage of students scored between 81 and 89?
d. Only 5% of the students taking the test scored higher than what grade?
Solution:
- We will declare a random variable X denoting the score that a student gets on a final exam. So,
X ~ N ( 69 , 9 )
- After defining our variable X follows a normal distribution. We can compute the probabilities as follows:
a) P ( X < 91 ) ?
- Compute the Z-score value as follows:
Z = (91 - 69) / 9 = 2.4444
- Now use the Z-score tables and look for z = 2.444:
P( X < 91 ) = P ( Z < 2.4444) = 0.9927
b) P ( 65 < X < 89 ) ?
- Compute the Z-score values as follows:
Z = (89 - 69) / 9 = 2.2.222
Z = (65 - 69) / 9 = -0.4444
- Now use the Z-score tables and look for z = 2.222 and Z = -0.4444:
P(65 < X < 89 ) = P ( -0.444< Z < 2.2222) = 0.6585
b) P ( 81 < X < 89 ) ?
- Compute the Z-score values as follows:
Z = (89 - 69) / 9 = 2.2.222
Z = (81 - 69) / 9 = 1.3333
- Now use the Z-score tables and look for z = 2.222 and Z = 1.333:
P(81 < X < 89 ) = P ( 1.333< Z < 2.2222) = 0.0781
c) P ( X > a ) = 0.05 , a?
- Compute the Z-score values as follows:
Z = (a - 69) / 9 = q
- Now use the Z-score tables and look for z value that corresponds to:
P( X > a ) = P ( Z > q ) = 0.05
- The corresponding Z-value is: q = 1.6444
Hence,
Z = (a - 69) / 9 = 1.644
a = 83.8
