A sample of students had a mean age of 35 years along with a standard deviation of 5 years. A student was randomly picked from a group of 200 students. Determine the probability that the age of the student turned out to be as given below:
i. Lying in between 35 and 40
ii. Lying in between 30 and 40
iii. Lying in between 25 and 30
iv. Lying beyond 45 yrs
v. Lying beyond 30 yrs
vi. Lying below 25 years
Solution
(i). The standardized value for 35 years
Z = (x - μ)/s = (35-35)/5 = 0
The standardized value for 40 years
Z = (x - μ)/s = (40- 35)/5 = 1
∴ the area between Z = 1 and Z = 0 is 0.3413. These values are checked from the general tables
The value from standard common curve tables
While z = 0, p = 0
And while z = 1, p = 0.3413
Now the area under this curve is the area between z = 0 and z = 1
= 0.3413 - 0 = 0.3413
∴ the probability age lying among 35 and 40 yrs is 0.3413
(ii). 30 and 40 years
Z = (x - μ)/s = (30 - 35)/5 = -5/5 = -1
Z = (x - μ)/s = (40 - 35)/5 = 5/5 = 1
∴ The area between Z = 1 and Z = -1 is = 0.3413 as lying on the positive side of zero + 0.3413 as lying on the negative side of zero
P = 0.6826
∴ the probability age lying between 30 and 40 yrs is 0.6826
(iii). 25 and 30 years
Z = (x - μ)/s = (25 - 35)/5 = -10/5 = -2
Z = (x - μ)/s = (30 - 35)/5 = -5/5 = -1
∴ the area between Z = -2 and Z = -1
Probability area corresponding to Z = -2
= 0.4772 (the z value to check from the tables is 2)
Probability area corresponding to Z = -1
= 0.3413 (the z value for this case is 1)
∴ the probability that the age lies between 25 and 30 yrs
= 0.4772 - 0.3413 the area under this curve
P = 0.1359
iv). P(beyond 45 years) is verified as given = P(x > 45)
Z = (x - μ)/s = (40 - 35)/5 = 10/5 = +2
Probability corresponding to Z = 2 = 0.4772 = probability of between 45 and 35
∴ P(Age > 45yrs) = 0.5000 - 0.4772
= 0.0228