involves running a short program. If past experience indicates that 40% of all students will make no programming errors, compute the (approximate) probability that in a class of 50 students a. At least 25 will make no errors [Hint: Normal approximation to the binomial] b. Between 15 and 25 (inclusive) will make no errors

(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 .

7

step-by-step explanation: