answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
schepotkina [342]
2 years ago
12

For each of the questions below, a histogram is described. Indicate in each case whether, in view of the Central Limit Theorem,

you can be confident that the histogram would look like approximately a bell-shaped (normal) curve, and give a brief explanation why (one sentence is probably sufficient). There are no data for these questions, so you will not need to use the computer to answer these questions.
a. A police department records the number of 911 calls made each day of the year, and the 365 values are plotted in a histogram.
b. The day before an election, fifty different polling organizations each sample 500 people and record the percentage who say they will vote for the Democratic candidate. The 50 values are plotted in a histogram. The fifty polling organizations also record the average age of the 500 people in their sample, and the 50 averages are plotted in a histogram.
c. One hundred batteries are tested, and the lifetimes of the batteries are plotted in a histogram.
d. Two hundred students in a statistics class each flip a coin 50 times and record the number of heads. The numbers of heads are plotted in a histogram. Two hundred students in a statistics each roll a die 40 times and record the sum of the numbers they got on the 40 rolls. They make a histogram of the 200 sums.
e. One thousand randomly chosen people report their annual salaries, and these salaries are plotted in a histogram.
Mathematics
1 answer:
denis-greek [22]2 years ago
6 0

Solution :

a). This histogram is not a bell shaped.

    Here we cannot use the theorem of the central limit as the histogram is being plotted by using 365 samples data values of the one sample.

b). This histogram may be bell shaped approximately since the sample size is large.

   Here there are 50 sample each having a size of 500 and there is 50 sample proportions. The histogram shows sampling distribution of the sample proportion. Thus the central limit theorem can be used.

The histogram is approximately bell shaped as there is 50 samples of each having 500 size and an average of 50. Thus the central limit theorem can be used.

c). The histogram is not a bell shaped.

   In this case central limit theorem cannot be used as the histogram is plotted by using 100 sample data values of one sample.

d). The histogram is a bell shaped since the sample size is large.

   In this case 200 samples  of each having a size of 50 and a data values of 50. The histogram here shows a sampling distribution of the sample proportion. Thus the central limit theorem can be used.

In the second also the histogram is of bell shaped as the sample size is large.

There are 200 samples of each sample having a size of 40 and they have 40 data values , i.e. sum of rolls.

e). It is not a bell shaped.

  The histogram here is plotted using 100 sample data values having one sample. Therefore we cannot use the central limit theorem.

You might be interested in
You have just opened a new dance club, Swing Haven, but are unsure of how high to set the cover charge (entrance fee). One week
aliya0001 [1]

Answer:

q = -6p + 104

Step-by-step explanation:

Linear function:

A linear function has the following format:

q = mp + b

In which m is the slope and b is the q-intercept.

One week you charged $4 per guest and averaged 80 guests per night. The next week you charged $10 per guest and averaged 44 guests per night.

This means that we have these following points: (4,80), (10,44).

Finding the slope:

With a pair of points, the slope is given by the change in q divided by the change in p.

Change in q: 44 - 80 = -36

Change in p: 10 - 4 = 6

Slope: m = \frac{-36}{6} = -6

So

q = -6p + b

Finding b:

We replace one of the points. Replacing (4,80).

q = -6p + b

80 = -6*4 + b

b = 104

So

q = -6p + 104

7 0
2 years ago
each volleyball team in a league needs 6 players 2 alternates, and a coach. How many teams can be formed with 288 peolple? show
mariarad [96]
You divide 288 by 6 and get the answer of 48
3 0
2 years ago
Read 2 more answers
Is the fraction 315/8 irrational
Debora [2.8K]

Answer:

315/8 GIVES 39.375.

But 35.375 Is actually a rational number,because it is terminating after 3 decimal places.

Hence its not a rational number

3 0
2 years ago
Read 2 more answers
Diego has 75% of $10. Noah has 25% of $30. Diego thinks he has more money than Noah, but Noah thinks they have an equal amount o
Fynjy0 [20]

Answer:

Noah

Step-by-step explanation:

75%*10=7.5

25%*30=7.5

The answers are both 7.5, which are equal

5 0
2 years ago
Read 2 more answers
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
2 years ago
Other questions:
  • Solve the Quadratic Equation by factoring. 3r^2-16r-7=5
    5·1 answer
  • In a basket, 5/9 of the fruits are apples and the rest are oranges. 3/10 of the apples are green in colour. There are 15 green a
    8·1 answer
  • Graph g(x), where f(x) = 4x − 2 and g(x) = f(x + 1)
    13·1 answer
  • Mrs. Harrison has a jar full of cookies.
    13·2 answers
  • A researcher claims that the mean cost of raising a child from birth to age 2 by husband­wife families in the U.S. is $13,960. A
    5·1 answer
  • A random survey of seventh and eighth grade students was conducted to find out how many hours are spent doing homework each nigh
    13·1 answer
  • What is the constant of proportionality in the equation y = StartFraction x over 9 EndFraction? 0 StartFraction 1 over 9 EndFrac
    5·2 answers
  • A computer had 2 gigabytes of data stored on it when Jackie bought it, and she is storing an additional 3.5 gigabytes per year,
    14·1 answer
  • A manufacturer produces three models of bicycles. The times (in hours) required for assembling, painting, and packaging each mod
    12·1 answer
  • A(n) _____ is any side of a polygon that shares a side with only one angle of a pair of angles.
    6·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!