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Montano1993 [528]
1 year ago
7

A researcher wants to know how homeowners in her state feel about home warranties. She obtains a list of 879 homeowners in her s

tate, selects 300 of them at random, and emails a questionnaire to them. 64 questionnaires are returned. Of these, 80% stated that they believe that the current home warranty coverage needs to be completely overhauled. Classify the following groups as the population or the sample for this particular study.
a. all homeowners in the researcher's state
b. The 879 homeowners whose names were obtained
c. The 64 homeowners who returned the questionnaire
d. The 80% who stated that the current home warranty coverage needs to be completely overhauled
Mathematics
1 answer:
TEA [102]1 year ago
7 0

Answer:

a. The population

b. The accessible population

c. The sample

d. The statistic

Step-by-step explanation:

a. All the homeowners in the researchers state = The target population

b. The 879 homeowners whose name were obtained = The sample frame (accessible population)

c. The 64 homeowners who returned the questionnaire = Research participants (the sample), the subjects

d. The 80% who stated that the current home warranty coverage needs to be completely overhauled = The statistic (subset or part of the sample)

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jeka57 [31]

-115

Using the arithmetic formula, your equation appears as so:

an = 15 + (27 - 1) - 5

15 being the first number in the sequence, 27 being the number you're trying to find, and -5 being the common difference.

This will give you the answer of -115.

Hope this helps!

8 0
2 years ago
Solve for d.
r-ruslan [8.4K]
5d + 2(2 - d) = 3(1 + d) + 1
5d + 2(2) - 2(d) = 3(1) + 3(d) + 1
5d + 4 - 2d = 3 + 3d + 1
5d - 2d + 4 = 3d + 3 + 1
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<u>-3d       -3d        </u>
       4 = 4
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4 0
2 years ago
Read 2 more answers
Suppose a scheduled flight must average at least 60% occupancy to be profitable. A sample of 120 flights on a particular route w
garik1379 [7]

Answer: see the graphic

Step-by-step explanation:

A. Type I error helps us to conclude that the flight is not profitable, when in fact it is profitable.

B. a = 0.05

C. Type II error does not show that the flight is profitable

8 0
2 years ago
The cafe where you work just ran out of coffee, and you are at the store to buy 112 pounds of coffee. You have put a can with 34
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Answer:

78

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The total you are buying is 112 pounds.  The can you have picked up is 34 pounds; this leaves

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2 years ago
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Three couples and two single individuals have been invited to an investment seminar and have agreed to attend. Suppose the proba
Maksim231197 [3]

Answer:

(a) Probability mass function

P(X=0) = 0.0602

P(X=1) = 0.0908

P(X=2) = 0.1704

P(X=3) = 0.2055

P(X=4) = 0.1285

P(X=5) = 0.1550

P(X=6) = 0.1427

P(X=7) = 0.0390

P(X=8) = 0.0147

NOTE: the sum of the probabilities gives 1.0068 for rounding errors. It can be divided by 1.0068 to get the adjusted values.

(b) Cumulative distribution function of X

F(X=0) = 0.0602

F(X=1) = 0.1510

F(X=2) = 0.3214

F(X=3) = 0.5269

F(X=4) = 0.6554

F(X=5) = 0.8104

F(X=6) = 0.9531

F(X=7) = 0.9921

F(X=8) = 1.0068

Step-by-step explanation:

Let X be the number of people who arrive late to the seminar, we can assess that X can take values from 0 (everybody on time) to 8 (everybody late).

<u>For X=0</u>

This happens when every couple and the singles are on time (ot).

P(X=0)=P(\#1=ot)*P(\#2=ot)*P(\#3=ot)*P(\#4=ot)*P(\#5=ot)\\\\P(X=0)=(1-0.43)^{5}=0.57^5= 0.0602

<u>For X=1</u>

This happens when only one single arrives late. It can be #4 or #5. As the probabilities are the same (P(#4=late)=P(#5=late)), we can multiply by 2 the former probability:

P(X=1) = P(\#4=late)+P(\#5=late)=2*P(\#4=late)\\\\P(X=1) = 2*P(\#1=ot)*P(\#2=ot)*P(\#3=ot)*P(\#4=late)*P(\#5=ot)\\\\P(X=1) = 2*0.57*0.57*0.57*0.43*0.57\\\\P(X=1) = 2*0.57^4*0.43=2*0.0454=0.0908

<u>For X=2</u>

This happens when

1) Only one of the three couples is late, and the others cooples and singles are on time.

2) When both singles are late , and the couples are on time.

P(X=2)=3*(P(\#1=l)*P(\#2=ot)*P(\#3=ot)*P(\#4=ot)*P(\#5=ot))+P(\#1=ot)*P(\#2=ot)*P(\#3=ot)*P(\#4=l)*P(\#5=l)\\\\P(X=2)=3*(0.43*0.57^4)+(0.43^2*0.57^3)=0.1362+0.0342=0.1704

<u>For X=3</u>

This happens when

1) Only one couple (3 posibilities) and one single are late (2 posibilities). This means there are 3*2=6 combinations of this.

P(X=3)=6*(P(\#1=l)*P(\#2=ot)*P(\#3=ot)*P(\#4=l)*P(\#5=ot))\\\\P(X=3)=6*(0.43^2*0.57^3)=6*0.342=0.2055

<u>For X=4</u>

This happens when

1) Only two couples are late. There are 3 combinations of these.

2) Only one couple and both singles are late. Only one combination of these situation.

P(X=4)=3*(P(\#1=l)*P(\#2=l)*P(\#3=ot)*P(\#4=ot)*P(\#5=ot))+P(\#1=l)*P(\#2=ot)*P(\#3=ot)*P(\#4=l)*P(\#5=l)\\\\P(X=4)=3*(0.43^2*0.57^3)+(0.43^3*0.57^2)\\\\P(X=4)=3*0.0342+ 0.0258=0.1027+0.0258=0.1285

<u>For X=5</u>

This happens when

1) Only two couples (3 combinations) and one single are late (2 combinations). There are 6 combinations.

P(X=6)=6*(P(\#1=l)*P(\#2=l)*P(\#3=ot)*P(\#4=l)*P(\#5=ot))\\\\P(X=6)=6*(0.43^3*0.57^2)=6*0.0258=0.1550

<u>For X=6</u>

This happens when

1) Only the three couples are late (1 combination)

2) Only two couples (3 combinations) and one single (2 combinations) are late

P(X=6)=P(\#1=l)*P(\#2=l)*P(\#3=l)*P(\#4=ot)*P(\#5=ot)+6*(P(\#1=l)*P(\#2=l)*P(\#3=ot)*P(\#4=l)*P(\#5=ot))\\\\P(X=6)=(0.43^3*0.57^2)+6*(0.43^4*0.57)\\\\P(X=6)=0.0258+6*0.0195=0.0258+0.1169=0.1427

<u>For X=7</u>

This happens when

1) Only one of the singles is on time (2 combinations)

P(X=7)=2*P(\#1=l)*P(\#2=l)*P(\#3=l)*P(\#4=l)*P(\#5=ot)\\\\P(X=7)=2*0.43^4*0.57=0.0390

<u>For X=8</u>

This happens when everybody is late

P(X=8)=P(\#1=l)*P(\#2=l)*P(\#3=l)*P(\#4=l)*P(\#5=l)\\\\P(X=8) = 0.43^5=0.0147

8 0
1 year ago
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