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3 years ago

What is the mean absolute deviation of the data set?

Mathematics

1 answer:

Ronch [10]3 years ago

5 0

Answer:

the mean absolute deviation of the data set is 2.

Step-by-step explanation:

To calculate the mean absolute deviation of a data set, we need to follow the following steps:

Step 1: Calculate the mean.

Step 2: Calculate how far away each data point is from the mean (always keeping a possitive distance).

Step 3: Calculate the mean of the deviations.

Step 1: The mean of the data set is:

Mean = (12 + 10 + 10 + 8 + 6+ 7 + 7 + 12)/8 = 72/8 = 9

Step 2:

{12 -9, 10-9, 10-9, 9-8, 9-6, 9-7, 9-7, 12-9} = {3, 1, 1, 1, 3, 2, 2, 3}

Step 3: (3 + 1 + 1 + 1 + 3 + 2 + 2 + 3)/8 = 2

In conclusion, the mean absolute deviation of the data set is 2. Which corresponds to the option B.

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Solve: 3a - 2.4 = 5.1 a = _____

Delicious77 [7]

Answer:

a=2.5

Step-by-step explanation:

3a-2.4=5.1 Add the like terms ( plus 2.4 cancels out -2.4)

+2.4 +2.4

3a=7.5 Now divide by 3

/3 /3

a=2.5

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3 years ago

What do you call identical twin sisters when both are ice skating champions

BigorU [14]

The answer for your question is Ice queen clones.

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3 years ago

Read 2 more answers

Rafael can choose plan a or Plan B for his long-distance charges. For each plan, cost (in dollars) depends on minutes used (Per

PilotLPTM [1.2K]

Answer:

a. Plan B; $4

b. 160 mins; Plan B

Step-by-step explanation:

a. Cost of Plan A for 80 minutes:

Find 80 on the x axis, and trave it up to to intercept the blue line (for Plan A). Check the y axis to see the value of y at this point. Thus:

f(80) = 8

This means Plan A will cost $8 for Rafael to 80 mins of long distance call per month.

Also, find the cost per month for 80 mins for Plan B. Use the same procedure as used in finding cost for plan A.

Plan B will cost $12.

Therefore, Plan B cost more.

Plan B cost $4 more than Plan A ($12 - $8 = $4)

b. Number of minutes that the two will cost the same is the number of minutes at the point where the two lines intercept = 160 minutes.

At 160 minutes, they both cost $16

The plan that will cost less if the time spent exceeds 160 minutes is Plan B.

4 0

3 years ago

6. Mike's family went to the YES Prep Volleyball game. Mike bought 1 ticket, 2 chips and a drink

KIM [24]

Answer:

x=5

y=1

z=2

Step-by-step explanation:

x+2y+z=9

3x+y+4z=24

x+3y+z=10

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3 years ago

⦁ In a simple random sample of 1219 US adults, 354 said that their favorite sport to watch is football. Construct a 95% confiden

fomenos

Answer:

95% confidence interval for the proportion of adults in the United States whose favorite sport to watch is football is [0.265 , 0.316].

Step-by-step explanation:

We are given that in a simple random sample of 1219 US adults, 354 said that their favorite sport to watch is football.

Firstly, the pivotal quantity for 95% confidence interval for the proportion of adults in the United States whose favorite sport to watch is football is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = proportion of adults in the United States whose favorite sport to watch is football in a sample of 1219 adults = $\frac{354}{1219}$

n = sample of US adults = 1291

p = population proportion of adults

<em>Here for constructing 95% confidence interval we have used One-sample z proportion statistics.</em>

So, 95% confidence interval for the population proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5%

significance level are -1.96 & 1.96}

P(-1.96 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ < ${\hat p-p}$ < $1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ ) = 0.95

P( $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ < p < $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ ) = 0.95

<u>95% confidence interval for p</u> = [ $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ , $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} }$ ]

= [ $\frac{354}{1219}-1.96 \times {\sqrt{\frac{\frac{354}{1219}(1-\frac{354}{1219})}{1219} }$ , $\frac{354}{1219}+1.96 \times {\sqrt{\frac{\frac{354}{1219}(1-\frac{354}{1219})}{1219} }$ ]

= [0.265 , 0.316]

Therefore, 95% confidence interval for the proportion of adults in the United States whose favorite sport to watch is football is [0.265 , 0.316].

5 0

3 years ago