Answer:
(a) P(X 25) = 0.0985
(b) P(15 X
25) = 0.8836
Step-by-step explanation:
We are given that past experience indicates that 40% of all students will make no programming errors. Also, a sample of class of 50 students is given.
Firstly, the above situation can be represented through binomial distribution, i.e.;
where, n = number of samples taken = 50
r = number of success
p = probability of success, i.e. 40%
Now, we can't calculate the required probability using binomial distribution because here n is very large(n > 30), so we will convert this distribution into normal distribution using continuity correction.
So, Let X = No. of making no programming errors
Mean of X, =
=
= 20
Standard deviation of X, =
=
= 3.5
So, X ~ N()
Now, the z score probability distribution is given by;
Z = ~ N(0,1)
(a) Probability that in a class of 50 students at least 25 will make no errors is given by = P(X 25) = P(X > 24.5) ---- using continuity correction
P(X > 24.5) = P( >
) = P(Z > 1.29) = 1 - P(Z
1.29)
= 1 - 0.90147 = 0.0985
(b) Probability that in a class of 50 students between 15 and 25 (inclusive) will make no errors = P(15 X
25) = P(X
25) - P(X < 15)
P(X 25) = P(X < 25.5) = P(
<
) = P(Z < 1.57) = 0.94179
P(X < 15) = P(X < 14.5) = P( <
) = P(Z < -1.57) = 1 - P(Z
1.57)
= 1 - 0.94179 = 0.05821
Therefore, P(15 X
25) = 0.94179 - 0.05821 = 0.8836 .