Question

Suppose 85% of all students taking a beginning programming course fail to get their first program to run on first submission. Consider a group of 7 such students,where each student's success is independent from the other and the chance each student fails on their first try is consistent. (Round answers to three decimal places.)
a. What is the probability exactly 6 fail on their first submissions?
b. What is the probability 6 or less fail on their first submissions?
c. What is the probability at least 6 fail on their first submissions?
d. How many students should be expected to fail?
e. What is the standard deviation?

Answer 1

Answer:

a) $P(X=6)=(7C6)(0.85)^6 (1-0.85)^{7-6}=0.396$

b) $P(X\leq 6)$

And we can use the complement rule and we got:

$P(X\leq 6)= 1- P(X>6) = 1-P(X = 7)$

$P(X=7)=(7C7)(0.85)^7 (1-0.85)^{7-7}=0.321$

And replacing we got:

$P(x \leq 6) = 1-0.321= 0.679$

c) $P(X\geq 6)= P(X=6) +P(X=7) = 0.396+0.321= 0.717$

d) $E(X) = 7*0.85= 5.95$

e) $\sigma = \sqrt{7*0.85*(1-0.85)}=0.945$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=7, p=0.85)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

Part a

We want this probability:

$P(X=6)$

And using the probability mas function we got:

$P(X=6)=(7C6)(0.85)^6 (1-0.85)^{7-6}=0.396$

Part b

We want this probability:

$P(X\leq 6)$

And we can use the complement rule and we got:

$P(X\leq 6)= 1- P(X>6) = 1-P(X = 7)$

$P(X=7)=(7C7)(0.85)^7 (1-0.85)^{7-7}=0.321$

And replacing we got:

$P(x \leq 6) = 1-0.321= 0.679$

Part c

We want this probability:

$P(X\geq 6)= P(X=6) +P(X=7) = 0.396+0.321= 0.717$

Part d

the expected value is given by:

$E(X) = 7*0.85= 5.95$

Part e

The standard deviation is given by:

$\sigma = \sqrt{7*0.85*(1-0.85)}=0.945$