Question

Letter C??

Data were collected from a survey given to graduating college seniors on the number of times they had changed majors. From that data, a probability distribution was constructed. The random variable X is defined as the number of times a graduating senior changed majors. It is shown below: P(X = ?) 0.18 0.276 0.228 0.193 0.068 0.038 0.013 0.003 0.001 a. What is the probability that a randomly selected student changed his or her major at least once? Preview 0.82 0.82 b. What is the probability that a randomly selected student changed his or her major at most twice? 0.684 Preview 0.684 c. Given that a randomly selected person did change majors, what is the probability that he or she changed majors more than three times? | Preview Box 1: Enter your answer as a number (like 5, -3, 2.2172) or as a calculation (like 5/3, 243, 5+4) Enter DNE for Does Not Exist, oo for Infinity Box 2: Enter your answer as a number (like 5, -3, 2.2172) or as a calculation (like 5/3, 243, 5+4) Enter DNE for Does Not Exist, oo for Infinity Box 3: Enter your answer as a number (like 5, -3, 2.2172) or as a calculation (like 5/3, 243, 5+4) Enter DNE for Does Not Exist, oo for Infinity

Answer 1

We are given that the random variable X which represents the number of times a graduating senior changed major.

a.)

The probability that a randomly selected student changed his or her major at least once.

It means x should be at least once $(x~\geq~1)$

Thus,

Plc > 1) = 1 - Pir <1)

$P(x~\geq~1)~=~1~-~P(x~=~0)$

$P(x~\geq~1)~=~1~-~0.18 ~=~\huge\boxed{0.82}$

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b.)

Probability x to be at most twice.

Pl.x < 2) = P(r = 0) + Plit = 1) + Plix = 2)

$P(x~\leq~2)~=0.18~+~0.276~+~0.228~=~\huge\boxed{0.684}$

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c.)

Given that randomly selected person did change majors.

It means that x = o would no be considere because it implies that a senior did change it's major 0 times or not changed.

We have to find the probability that he or she changed more than three times i.e.

Since, we have given statemet, thus it is a case of conditional probabiity.

So, we have to find:

$\LARGE \mathrm{P(x~>~3| not ~x = 0)~=~\frac{P(x~>~3~\cap~not~x = 0)}{P(not ~x ~=~0)}~=~\frac{P(x~>~3~and~not~x = 0)}{P(not ~x ~=~0)}}$

NOTE: (not x= 0) means x can take value other than 0 which is 1, 2, 3, 4, 5, 6, 7, 8

and (x > 3) means x can take values 4, 5, 6, 7, 8

intersection $\LARGE \cap$ refers to "and"

Thus, x can overall take values 4, 5, 6, 7, 8

Then,

$\LARGE \mathrm{P(x~>~3| not ~x = 0)~=~\frac{P(x~>~3~\cap~not~x = 0)}{P(not ~x ~=~0)}}$

$\LARGE \mathrm{P(x~>~3| not ~x = 0)~=~\frac{P(x~=~4,5,6,7,8)}{P(not ~x ~=~0)}}$

This means a senior did change its major either 4 times, or 5 times or 6 times or 7 times or 8 times.

and we know that or refers to union $\LARGE \cup$ .

$\LARGE \mathrm{P(x~>~3| not ~x = 0)~=~\frac{P(x~=~4)\cup~P(x~=~5) \cup~P(x~=~6)\cup P(x~=~7)\cup P(x~=~8)}{1~-~P(x ~=~0)}}$

$\LARGE \mathrm{P(x~>~3| not ~x = 0)~=~\frac{ 0.068+0.038+0.013+0.003+0.001}{1~-~0.18}}$

$\LARGE \mathrm{P(x~>~3| not ~x = 0)~=~\frac{ 0.123}{0.82}~=~\huge\boxed{0.15}}$