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Solved:The first assignment in a statistical computing class involves running a short program. If past experience indicates
ByDr. Raju Chaudhari
May 31, 2021 Binomial Distribution, normal distribution, probability distribution
The first assignment in a statistical computing class 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
Here $X$ denote the number of students who make no errors.
$p$ be the probability of making no programming errors.
Given that $p=0.4$ and $n =50$. Thus $X\sim B(50, 0.4)$.
Mean = $\mu= n*p = 50 \times 0.4 = 20$.
sd = $\sigma= \sqrt{n*p*(1-p)} = \sqrt{50 \times 0.4 \times (1- 0.4)}=3.4641$
(a) The (approximate) probability that in a class of 50 students
at least 25 will make no errors is
$$ \begin{aligned} P(X\geq 25) & = 1- P(X < 25)\\ &= 1- P(X < 24.5)\\ &\quad \quad (\text{ using continuity correction})\\ &= 1-P\bigg(\frac{X-\mu}{\sigma} < \frac{24.5- 20}{3.4641}\bigg)\\ & =1-P(Z < 1.299)\\ & = 1-0.903\\ & = 0.097 \end{aligned} $$
(b) The (approximate) probability that in a class of 50 students between 15 and 25 (inclusive) will make no errors is
$$ \begin{aligned} P(15\leq X\leq 25) &= P\bigg(\frac{15- 20}{3.4641}\leq \frac{X-\mu}{\sigma} < \frac{25- 20}{3.4641}\bigg)\\ &= P(-1.4434 < Z < 1.4434)\\ & =P(Z < 1.4434)-P(Z < -1.4434)\\ & = 0.9255-0.0745\\ & = 0.851 \end{aligned} $$
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