22.03.2022

Clarkson University surveyed alumni to learn more about what they think of Clarkson. One part of the survey asked respondents to indicate whether their overall experience at Clarkson fell short of expectations, met expectations, or surpassed expectations. The results showed that 5% of the respondents did not provide a response, 26% said that their experience fell short of expectations, and 65% of the respondents said that their experience met expectations. (a) If we chose an alumnus at random, what is the probability that the alumnus would say their experience surpassed expectations

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24.06.2023, solved by verified expert
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0.04

Step-by-step explanation:

We are given that

The probability  that the alumnus did not provide response=5%

The probability  that the alumnus did not provide response=Clarkson University surveyed alumni to learn, №17887446, 22.03.2022 07:48

The probability that the alumnus would say their experience  fell short of expectations=26%=0.26

The probability that the alumnus would say their experience met expectations=65%=0.65

We have to find the  probability that the alumnus would say their experience surpassed expectations.

We know that

Sum of all probabilities =1

Using the property

The  probability that the alumnus would say their experience surpassed expectations=1-(0.05+0.26+0.65)

The  probability that the alumnus would say their experience surpassed expectation=0.04

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Mathematics
Step-by-step answer
P Answered by PhD

(a) The probability that a randomly selected alumnus would say their experience surpassed expectations is 0.05.

(b) The probability that a randomly selected alumnus would say their experience met or surpassed expectations is 0.67.

Step-by-step explanation:

Let's denote the events as follows:

A = Fell short of expectations

B = Met expectations

C = Surpassed expectations

N = no response

Given:

P (N) = 0.04

P (A) = 0.26

P (B) = 0.65

(a)

Compute the probability that a randomly selected alumnus would say their experience surpassed expectations as follows:

P(C) = 1 - [P(A) + P(B) + P(N)]\\= 1 - [0.26 + 0.65 + 0.04]\\= 1 - 0.95\\= 0.05

Thus, the probability that a randomly selected alumnus would say their experience surpassed expectations is 0.05.

(b)

The response of all individuals are independent.

Compute the probability that a randomly selected alumnus would say their experience met or surpassed expectations as follows:

P(B\cup C) = P(B)+P(C)-P(B\cap C)\\=P(B)+P(C)-P(B)\times P(C)\\= 0.65 + 0.05 - (0.65\times0.05)\\=0.6675\\\approx0.67

Thus, the probability that a randomly selected alumnus would say their experience met or surpassed expectations is 0.67.

Mathematics
Step-by-step answer
P Answered by PhD

0.04

Step-by-step explanation:

We are given that

The probability  that the alumnus did not provide response=5%

The probability  that the alumnus did not provide response=\frac{5}{100}=0.05

The probability that the alumnus would say their experience  fell short of expectations=26%=0.26

The probability that the alumnus would say their experience met expectations=65%=0.65

We have to find the  probability that the alumnus would say their experience surpassed expectations.

We know that

Sum of all probabilities =1

Using the property

The  probability that the alumnus would say their experience surpassed expectations=1-(0.05+0.26+0.65)

The  probability that the alumnus would say their experience surpassed expectation=0.04

Mathematics
Step-by-step answer
P Answered by PhD

Step-by-step explanation:

The question is incomplete. The complete question is:

Clarkson University surveyed alumni to learn more about what they think of Clarkson. One part of the survey asked respondents to indicate whether their overall experience at Clarkson fell short of expectations, met expectations, or surpassed expectations. The results showed that 4% of respondents did not provide a response, 26% said that their experience fell short of expectations, 65% of the respondents said that their experience met expectations (Clarkson Magazine, Summer, 2001). If we chose an alumnus at random, what is the probability that the alumnus would say their experience surpassed expectations? If we chose an alumnus at random, what is the probability that the alumnus would say their experience met or surpassed expectations?

Solution:

Probability = number of favorable outcomes/number of total outcomes

From the information given,

The probability that respondents did not provide a response, P(A) is 4/100 = 0.04

The probability that a respondent said that their experience fell short of expectations, P(B) is 26/100 = 0.26

The probability that a respondent said that their experience met expectations, P(C) is 65/100 = 0.65

A) Adding all the probabilities, it becomes 0.04 + 0.26 + 0.65 = 0.95

Therefore, the probability,P(D) that a respondent said that their experience surpassed expectations is 1 - 0.95 = 0.05

B) The event of a randomly chosen respondent saying that their experience met expectations and that their experience surpassed expectations are mutually exclusive because they cannot occur together. It means that P(C) × P(D) = 0

Therefore, the probability of P(C) or P(D) is 0.65 + 0.05 = 0.7

Mathematics
Step-by-step answer
P Answered by PhD

SI=(P*R*T)/100

P=2000

R=1.5

T=6

SI=(2000*1.5*6)/100

=(2000*9)/100

=180

Neil will earn interest of 180

Mathematics
Step-by-step answer
P Answered by PhD

For every 8 cars there are 7 trucks

Therefore,

Cars:Truck=8:7

Answer is B)8:7

Mathematics
Step-by-step answer
P Answered by PhD

y=2x+15

where y=Value of coin

x=Age in years

Value of coin after 19 years=2*19+15

=$53

Therefore, Value after 19 years=$53

Mathematics
Step-by-step answer
P Answered by PhD

F=ma

where F=force

m=mass

a=acceleration

Here,

F=4300

a=3.3m/s2

m=F/a

    =4300/3.3

    =1303.03kg

Mathematics
Step-by-step answer
P Answered by PhD

Salesperson will make 6% of 1800

=(6/100)*1800

=108

Salesperson will make $108 in $1800 sales

Mathematics
Step-by-step answer
P Answered by PhD

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