Question

Thefirstassignmentinastatisticalcomputingclass 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

Answer 1

Answer:

(a) P(X $\geq$ 25) = 0.0985

(b) P(15 $\leq$ X $\leq$ 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.;

$P(X=r) = \binom{n}{r} p^{r} (1-p)^{2} ;x=0,1,2,3,....$

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, $\mu$ = $n \times p$ = $50 \times 0.40$ = 20

Standard deviation of X, $\sigma$ = $\sqrt{np(1-p)}$ = $\sqrt{50 \times 0.40 \times (1-0.40)}$ = 3.5

So, X ~ N($\mu = 20, \sigma^{2} = 3.5^{2}$)

Now, the z score probability distribution is given by;

                 Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

(a) Probability that in a class of 50 students at least 25 will make no errors is given by = P(X $\geq$ 25) = P(X > 24.5)  ---- using continuity correction

   P(X > 24.5) = P( $\frac{X-\mu}{\sigma}$ > $\frac{24.5-20}{3.5}$ ) = P(Z > 1.29) = 1 - P(Z $\leq$ 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 $\leq$ X $\leq$ 25) = P(X $\leq$ 25) - P(X < 15)

P(X $\leq$ 25) = P(X < 25.5) = P( $\frac{X-\mu}{\sigma}$ < $\frac{25.5-20}{3.5}$ ) = P(Z < 1.57) = 0.94179

P(X < 15) = P(X < 14.5) = P( $\frac{X-\mu}{\sigma}$ < $\frac{14.5-20}{3.5}$ ) = P(Z < -1.57) = 1 - P(Z $\leq$ 1.57)

                                                                   = 1 - 0.94179  = 0.05821

Therefore, P(15 $\leq$ X $\leq$ 25) = 0.94179 - 0.05821 = 0.8836 .