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A random sample of 25 statistics examinations was taken. the average score in the sample was 76 with a standard deviation of 12. assuming the scores are normally distributed, the 99% confidence interval for the population average examination score is:

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Since the sample size is less than 30, therefore we use
the t statistic.

Let us define the given variables:

N = sample size = 25

X = average score = 76

s = standard deviation = 12

99% Confidence interval

Degrees of freedom = n – 1 = 24

 

The formula for confidence interval is given as:

CI = X ± t * s / sqrt N

 

using the standard distribution table, the t value for DF
= 24 and 99% CI is:

t = 2.492

 

Therefore calculating the CI using the known values:

CI = 76 ± 2.492 * 12 / sqrt 25

CI = 76 ± 5.98

CI = 70.02, 81.98

 

Answer: The average score ranges from 70 to 82.

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Suppose that people’s heights (in centimeters) are normally distributed with a mean of 170 and a standard deviation of 5. We find the heights of 50 people.

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We have
mean=mu=170
standard deviation=sigma=5

can now calculate the Zmin and Zmax using Z=(X-mean)/standard deviation
Zmin=(165-170)/5=-1
Zmax=(175-170)/5=+1
From normal probability tables, 
P(z<Zmin)=P(z<-1)=0.15866
P(z<Zmax)=P(z<+1)=0.84134
P(165<x<175)=P(Zmin<z<Zmax)=0.84134-0.15866= 0.68269

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What price do farmers get for the peach crops? in the third week of June, a random sample of 40 farming regions gave a sample mean of $6.88 per basket. assume that the standard deviation is known to be $1.92 per basket. find a 90% confidence interval for the population mean price per basket that farmers in this region get for their peach crop

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Given:
Sample size, n = 40
Sample mean, xb = $6.88
Population std. deviation, σ = $1.92 (known)
Confidence interval = 90%

Assume normal distribution for the population.
The confidence interval is
(xb + 1.645*(σ/√n), xb – 1.645*(σ/√n)
= (6.88 + (1.645*1.92)/√40, 6.88 – (1.645*1.92)/√40)
= (7.38, 6.38)

Answer: The 90% confidence interval is (7.38, 6.38)

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Consider college officials in admissions registration, counseling, financial aid campus ministry, food services, and so on. How much money do these people make each year? Suppose you read in your local newspaper that 45 officials in student services earn an average of $50,340 each year. Assume that the standard deviation is $10,780 for salaries of college officials and student services. Find a 90% confidence interval for the population mean salaries of such personnel. Round your answer to the nearest dollar and don;t forget to use the $ sign.

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Since sample size is > 40, we use the z-score
in calculating for the confidence interval.

The formula is given as:

Confidence Interval = X ± z * σ / sqrt (n)

Where,

X = mean = $50,340

z = z-score which is taken from standard distribution
tables at 90% confidence interval = 1.645

σ
= standard deviation = $10,780

n = sample size = 45

Substituting to the equation:

Confidence Interval = 50,340 ± 1.645 * 10,780 / sqrt (45)

Confidence Interval = 50,340 ± 1,607

Confidence Interval = $48,733 to $51,947

Therefore the salary range of the personnel is $48,733 to $51,947.

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The weights of certain machine components are normally distributed with a mean of 8.5 g and a standard deviation of 0.09 g. Find the two weights that separate the top 3% and the bottom 3%. These weights could serve as limits used to identify which components should be rejected. Round to the nearest hundredth of a gram.

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To solve this problem,lets say that

X = the weight of the machine components. 
X is normally distributed with mean=8.5 and sd=0.09

We need to find x1 and x2 such that 
P(X<x1)=0.03 and P(X>x2)=0.03 

Standardizing:

P( Z< (x1 – 8.5)/0.09 ) =0.03
P(Z > (x2 – 8.5)/0.09 ) =0.03. 

From the Z standard table, we can see that approximately P
= 0.03 is achieved when Z equals to:

z = -1.88          and      z= 1.88

Therefore,

P(Z<-1.88)=0.03 and P(Z>1.88)=0.03 

So, 

(x1 – 8.5)/0.09 = -1.88 and 
(x2 – 8.5)/0.09 =1.88 

Solving for x1 and x2: 

x1=-1.88(0.09) + 8.5   and
x2=1.88(0.09) + 8.5

Which yields:


x1 = 8.33 g

x2 = 8.67 g

Answer: The bottom 3 is separated by the weight
8.33 g and the top 3 by the weight 8.67 g.

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CAN YOU PLEASE HELP ME WITH THIS PROBLEM? This is due by MIDNIGHT 2). Manufacturer A produces hammers that are normally distributed with a mean weight of 4.6 lb and a standard deviation of 0.8 lb. Manufacturer B produces hammers that are normally distributed with a mean weight of 6.3 lb and a standard deviation of 1.4 lb. (a) What percentage of Manufacturer A’s hammers will weigh less than 5 lb? (b) What percentage of Manufacturer B’s hammers will weigh less than 5 lb? (c) Which manufacturer is more likely to produce a hammer weighing exactly 5.2 lb? Explain.

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Part (a)
Manufacturer A;
Mean, μ = 4.6
Std. deviation, σ = 0.8

For the random variable x = 5, the z-score is
z = (x – μ)/σ = (5 – 4.6)/0.8 = 0.5
From standard tables,
P(x<5) = P(z<0.5) = 0.69 = 69%

Answer:  
69% will weigh less than 5 lb.

Part b)
Manufacturer B
μ = 6.3
σ = 1.4

For x = 5, z = (5 – 6.3)/1.4 = -0.9286
From standard tables,
P(x<5) = P(z<-0.9286) = 0.1766 = 17.7%

Answer:
About 18% will weigh less than 5 lb.

Part (c)
x = 5.2 lb

Manufacturer A:
z = (5.2-4.6)/0.8 =  0.75
From tables,
P(x=5.2) = 0.773 = 77.3%

Manufacturer B:
z = (5.2-6.3)/1.4 = -0.7857
P(x=5.2) = 0.216 = 21.6%

Answer:
Manufacturer A is more likely to produce a 5.2 lb hammer because its probability is higher.

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The body temperatures of adults are normally distributed with a mean of 98.6° f and a standard deviation of 0.60° f. if 36 adults are randomly selected, find the probability that their mean body temperature is greater than 98.4° f. 0.0228 0.8188 0.9360 0.9772

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Here the population standard deviation is 0.60 degree F.  If a sample of 36 adults is randomly selected, that results in a sample standard deviation of 0.60 degree F divided by the square root of 36:  0.10 degree F.

The probability in question is the area under the standard normal probability distribution between 98.4 degree F and infinity, and intuitively you can detect that this will be more than 0.5 (corresponding to 50%).

Convert 98.4 degrees F to a z-score, using the sample standard deviation (0.10 degree F).  That z score is 
       98.4-98.6
z = ————–   =  -0.20/0.10 = -2
           0.10

We need to determine the area under the standard normal curve to the right of z=-2.  Use a table of z-scores to do this, or use your calculator’s built-in probability functions.  My result is 98.21% (corresponding to an area of 0.9821).

With my calculator I can find this probability using the following command:

normalcdf(-2,100000,0.10).

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Given that the random variable x is normally distributed with a mean of 80 and a standard deviation of 10, p(85 < x < 90) is

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Given that the random variable x is normally distributed with a mean of 80 and a standard deviation of 10, p(85 < x < 90) is

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A normal distribution of data has a mean of 15 and a standard deviation of 4. How many standard deviations from the mean is 25? 0.16 0.4 2.5 6.25

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Answered by answersmine AT 22/10/2019 – 05:19 AM

This is the concept of probability, to get the number of standard deviations that 25 is from the mean, we calculate the z-score given by:
Z=(X-mean)/s.d
where;
x=25
mean=15
s.d=4
hence;
z=(25-15)/4=2.5
The answer is 2.5

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Find the sample standard deviation and the population standard deviation of the data set. 70, 58, 70, 37, 58, 47, 58, 76, 77, 67, 66, 77, 33, 74, 57

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Employee                                 Mary      Zoe         Greg         Ann           Tom

Cumulative Pay                       $6,800   $10,500  $8,400    $66,000   $4,700

Pay subject to FICA S.S.         $421.60  $651.00  $520.80 $4092.00 $291.40
6.2%, (First $118,000)

Pay subject to FICA Medicare $98.60 $152.25    $121.80    $957.00    $68.15
1.45% of gross

Pay subject to FUTA Taxes      $40.80  $63.00     $50.40    $396.00  $28.20
0.6%

Pay subject to SUTA Taxes   $367.20  $567.00  $453.60  $3564.00 $253.80
5.4% (First $7000)

Totals                                     $928.20 $1,433.25 $1,146.60 $9,009.00 $641.55

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Imagine tossing a coin: What are your chances for tossing a head? What are your chances for tossing a tail? A coin is tossed 10 times: How many times do you expect to get heads? How many times do you expect to get tails? Using this data for the 10 coin tosses, calculate and record the deviation observed from what you expected using the following formula: The more the experimental results deviate from the expected results, the more the deviation value will approach the value of 1.0. As your results get closer to the expected results, the deviation is smaller and nears the value of 0.0. Interpret the meaning of the deviation value you obtained. A coin is tossed 100 times: How many times do you expect to get heads? How many times do you expect to get tails? Using this data for 100 coin tosses: Calculate the deviation for the 100 tosses. The 100 coin tosses are repeated. The results are added to the results of the first 100 coin tosses. How many times do you expect to get heads out of the 200 tosses? How many times do you expect to get tails out of the 200 tosses? Using this data for the 200 coin tosses: What is the deviation for the 200 tosses? How does increasing the total number of coin tosses from 10 to 100 affect the deviation? How does increasing the total number of tosses from 100 to 200 (or more) affect the deviation? What two important probability principles were established in this exercise? With two coins, both coins are tossed 100 times. How many times do you expect to get 2 heads? How many times do you expect to get 2 tails? How many times do you expect to get one head and one tail? The percent of occurrence is the obtained results divided by the total tosses and multiplied by 100. Using this data for the two coins being tossed 100 times. Calculate the percent occurrence for each combination: What is the percent of occurrence for two heads? What is the percent of occurrence for two tails? What is the percent of occurrence for one head and one tail?

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Answer:

The correct answer is a. Amy: wildlife in or near water = turtles, crayfish, goldfish non-water wildlife = fox, deer, bobcat.

Explanation:

According to the question students were asked to contrast their observations based on the location so according to the location, animals should be differentiated into water animals and land animals.

So turtles, crayfish, goldfish are the wildlife found in or near the water and fox, deer, bobcat are non-water animals according to their location. Therefore the correct answer is a. Amy: wildlife in or near water = turtles, crayfish, goldfish non-water wildlife = fox, deer, bobcat.

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Suppose a normal distribution has a mean of 98 and a standard deviation of 6. What is P(x less than or equal to 92)? A. 0.84 B. 0.16 C. 0.025 D. 0.975

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The basis to respond this question are:

1) Perpedicular lines form a 90° angle between them.

2) The product of the slopes of two any perpendicular lines is – 1.

So, from that basic knowledge you can analyze each option:

a.Lines s and t have slopes that are opposite reciprocals.

TRUE. Tha comes the number 2 basic condition for the perpendicular lines.

slope_1 * slope_2 = – 1 => slope_1 = – 1 / slope_2, which is what opposite reciprocals means.

b.Lines s and t have the same slope.

FALSE. We have already stated the the slopes are opposite reciprocals.

c.The product of the slopes of s and t is equal to -1

TRUE: that is one of the basic statements that you need to know and handle.

d.The lines have the same steepness.

FALSE: the slope is a measure of steepness, so they have different steepness.

e.The lines have different y intercepts.

FALSE: the y intercepts may be equal or different. For example y = x + 2 and y = -x + 2 are perpendicular and both have the same y intercept, 2.

f.The lines never intersect.

FALSE: perpendicular lines always intersept (in a 90° angle).

g.The intersection of s and t forms right angle.

TRUE: right angle = 90°.

h.If the slope of s is 6, the slope of t is -6

FALSE. – 6 is not the opposite reciprocal of 6. The opposite reciprocal of 6 is – 1/6.

So, the right choices are a, c and g.

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Sat scores are distributed with a mean of 1,500 and a standard deviation of 300. you are interested in estimating the average sat score of first year students at your college. if you would like to limit the margin of error of your 95% confidence interval to 25 points, how many students should you sample?

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Sat scores are distributed with a mean of 1,500 and a standard deviation of 300. you are interested in estimating the average sat score of first year students at your college. if you would like to limit the margin of error of your 95% confidence interval to 25 points, how many students should you sample?

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Which statements are true about mean absolute deviation? Select all that apply. The mean absolute deviation is impacted by outliers. The mean absolute deviation is a measure of center of the data. The mean absolute deviation is a measure of variation, or spread, of the data. The lower the mean absolute deviation is, the more spread out the data points are from the mean. The higher the mean absolute deviation is, the more spread out the data points are from the mean.

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The basis to respond this question are:

1) Perpedicular lines form a 90° angle between them.

2) The product of the slopes of two any perpendicular lines is – 1.

So, from that basic knowledge you can analyze each option:

a.Lines s and t have slopes that are opposite reciprocals.

TRUE. Tha comes the number 2 basic condition for the perpendicular lines.

slope_1 * slope_2 = – 1 => slope_1 = – 1 / slope_2, which is what opposite reciprocals means.

b.Lines s and t have the same slope.

FALSE. We have already stated the the slopes are opposite reciprocals.

c.The product of the slopes of s and t is equal to -1

TRUE: that is one of the basic statements that you need to know and handle.

d.The lines have the same steepness.

FALSE: the slope is a measure of steepness, so they have different steepness.

e.The lines have different y intercepts.

FALSE: the y intercepts may be equal or different. For example y = x + 2 and y = -x + 2 are perpendicular and both have the same y intercept, 2.

f.The lines never intersect.

FALSE: perpendicular lines always intersept (in a 90° angle).

g.The intersection of s and t forms right angle.

TRUE: right angle = 90°.

h.If the slope of s is 6, the slope of t is -6

FALSE. – 6 is not the opposite reciprocal of 6. The opposite reciprocal of 6 is – 1/6.

So, the right choices are a, c and g.

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Mark owns Siberian Husky sled dogs. He knows from data collected over the years that the weight of the dogs is a normal distribution. They have a mean weight of 52.5 lbs and a standard deviation of 2.4 lbs. What percentage of his dogs would you expect to have a weight between 47.7 lbs and 54.9 lbs?

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Employee                                 Mary      Zoe         Greg         Ann           Tom

Cumulative Pay                       $6,800   $10,500  $8,400    $66,000   $4,700

Pay subject to FICA S.S.         $421.60  $651.00  $520.80 $4092.00 $291.40
6.2%, (First $118,000)

Pay subject to FICA Medicare $98.60 $152.25    $121.80    $957.00    $68.15
1.45% of gross

Pay subject to FUTA Taxes      $40.80  $63.00     $50.40    $396.00  $28.20
0.6%

Pay subject to SUTA Taxes   $367.20  $567.00  $453.60  $3564.00 $253.80
5.4% (First $7000)

Totals                                     $928.20 $1,433.25 $1,146.60 $9,009.00 $641.55

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PLEASE RESPOND QUICK WITH THE CORRECT ANSWER OR YOU WILL GET REPORTED (08.03)Magdeline wants to know if the number of words on a page in her computer science book is generally more than the number of words on a page in her math book. She takes a random sample of 25 pages in each book and then calculates the mean, median, and mean absolute deviation for the 25 samples of each book. Book Mean Median Mean Absolute Deviation Computer science 48.7 40 9.4 Math 34.2 45 1.9 She claims that because the mean number of words on each page in the computer science book is greater than the mean number of words on each page in the math book, the computer science book has more words per page. Based on the data, is this a valid inference?

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Employee                                 Mary      Zoe         Greg         Ann           Tom

Cumulative Pay                       $6,800   $10,500  $8,400    $66,000   $4,700

Pay subject to FICA S.S.         $421.60  $651.00  $520.80 $4092.00 $291.40
6.2%, (First $118,000)

Pay subject to FICA Medicare $98.60 $152.25    $121.80    $957.00    $68.15
1.45% of gross

Pay subject to FUTA Taxes      $40.80  $63.00     $50.40    $396.00  $28.20
0.6%

Pay subject to SUTA Taxes   $367.20  $567.00  $453.60  $3564.00 $253.80
5.4% (First $7000)

Totals                                     $928.20 $1,433.25 $1,146.60 $9,009.00 $641.55

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4. Find the mean absolute deviation (MAD) for the following data: 77, 74, 81, 83, 78, 84, 79, 82, 75, 82, 79, 79

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4. Find the mean absolute deviation (MAD) for the following data: 77, 74, 81, 83, 78, 84, 79, 82, 75, 82, 79, 79

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The cost of a daily newspaper varies from city to city. however, the variation among prices remains steady with a standard deviation of 20?. a study was done to test the claim that the mean cost of a daily newspaper is $1.00. ten costs yield a mean cost of 96? with a standard deviation of 18?. do the data support the claim at the 1% level?

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127=7 (mod n) means when 127 is divided by n, the division leaves a remainder of 7.

The statement is equivalent to
120=0 (mod n), meaning that n divides 120.

All divisors of 120 will satisfy the statement because 120 divided by a divisor (factor) will leave a remainder of 0.

Factors of 120 are:
n={1,2,3,4,5,6,8,10,12,15,20,24,30,40,60,120}, |n|=16.
You can count how many such values of n there are, and try to check that each one satisfies 127=7 mod n.

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It has been reported that 63% of adults aged 65 and older got their flu shots last year. in a random sample of 300 adults aged 65 and over, find the mean, the variance, and standard deviation for the number who got their flu shots.

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It has been reported that 63% of adults aged 65 and older got their flu shots last year. in a random sample of 300 adults aged 65 and over, find the mean, the variance, and standard deviation for the number who got their flu shots.

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The mean life of pairs of shoes is 40 months with a standard deviation of 8 months. if the life of the shoes is normally distributed, how many pairs of shoes out of one million will need replacement before 30 months? round your answer to the nearest integer.

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You need to determine the number of ways in which 30 competitors from 50 can qualify. First, you have to realize that the order is irrelevant, that is: it is the same competitor_1, competitor _2, competitor _3 than competitor_3, competitor_2, competitor_1, or any combination of those three competitors.

So, the number of ways is which 30 competitors from 50 can qualify is given by the formula of combinations, which is:

C (m,n) = m! / (n! * (m -n)! )

=> C (50,30) = 50! / (30! (50 – 30)! ) = (50!) / [30! (50 – 30)!] = 50! / [30! 20!] =

 = 47,129,212,243,960 different ways the qualifiying round of 30 competitors can be selected from the 50 competitors.

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A population, with an unknown distribution, has a mean of 80 and a standard deviation of 7. for a sample of 49, the probability that the sample mean will be larger than 82 is

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The answer must take into account that the order is irrelevant, that is that it is the same J, Q, K that Q, K, J, and K, J, Q and all the variations of those the three cards.

The number of ways you can draw 50 cards from 52 is 52*51*50*49*48*47*…4*3 (it ends in 3).
,
But the number of ways that those 50 cards form the same set repeats is 50! = 50*49*49*47*….3*2*1

So, the answer is (52*51*50*49*48*….*3) / (50*49*48*…*3*2*1) =  (52*51) / 2 = 1,326.

Note that you obtain that same result when you use the formula for combinations of 50 cards taken from a set of 52 cards:

C(52,50) = 52! / [(50)! (52-50)!] = (52*51*50!) / [50! * 2!] = (52*51) / (2) = 1,326.

Answer: 1,326

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Find the​ mean, variance, and standard deviation of the binomial distribution with the given values of n and p. n equals 123n=123​, p equals 0.85

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Answer:

False

Step-by-step explanation:

Suposse that we are given a function f(x) and a constant value h.

1. Case:

If we take the function g(x)=f(x)+h, then the graph of the function g(x) will be the graph of the funcion f(x) moved up or down.

2.Case:

If we take the function g(x)=hf(x), then the graph of the function g(x) will be the graph of the function f(x) just taller or shorter.

3.Case:

If we take the function g(x)=f(x-h), then the graph of the function g(x) will be the graph of the fuction f(x) moved horizontally.

4. Case:

If we take the function g(x)=f(hx), then the graph of the function g(x) will be tha graph of the function f(x) wither or thiner.

For example:

If we take f(x)=sin(x) and h=2. Then, if we take g(x)=sin(2x) then f(0)=g(0)=0, which means that the graph of the functiction is not moved up or down. However, f(π/2)=sin(π/2)=1 and g(π/2)=sin(π)=0 which gives us a hint that the graph of the function became thiner.

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The manager of a warehouse would like to know how many errors are made when a product’s serial number is read by a bar-code reader. Six samples are collected of the number of scanning errors: 36, 14, 21, 39, 11, and 2 errors, per 1,000 scans each. What is the mean and standard deviation for these six samples?

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Answer:

Number of copy machines must be made to minimize the unit cost=160.

Step-by-step explanation:

We are given that the unit cost function C ( the cost in dollars to make each copy machine)

If machines are made =x

Then the unit cost function is given by

C(x)=0.8x^2-256x+25939

We have to find the number of copy machines for  minimize the unit  cost

C(x)=0.8x^2-256x+25939

Differentiate with respect to x

Then we get

frac{mathrm{d}C}{mathrm{d}x}=1.6x-256 ……(equation I)

To find the value of x then we susbtitute frac{mathrm{d}C}{mathrm{d}x} is equal to zero

frac{mathrm{d}C}{mathrm{d}x}=0

1.6x-256=0

1.6x=256

x=frac{256}{1.6}

By using division property of equality

x= 160

Again differentiate the equation I with respect to x then we get

frac{mathrm{d}^2 C}{mathrm{d}^2 x}=1.6 >0

Hence, the unit cost is minimize for x=160

Therefore, the number of copy machines must be made to minimize the unit cost =160

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Brian collects a set of 20 data representing the lengths of worms he found in the garden. The variance of measurments is 36. What is the standard deviation from the mean?

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Ur points are set up like this (x,y)….so ur x axis number, then ur y axis number…so 2 of the points that are on ur line are (0,12000) and (120,0)
u do not subtract 2000 by 4000…if u wanted to use those points, they would be (80,4000) and (100,2000)

slope (rate of change) = (y2 – y1) / (x2 – x1)
(0,12000) …x1 = 0 and y1 = 12000
 (120,0) ……x2 = 120 and y2 = 0
slope = (0 – 12,000) / (120 – 0) = -12,000/120 = -100 <== ur slope

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How can you use the mean absolute deviation of two data sets to compare them?

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The scatter graph of the data is shown in the first picture below

The ‘line of best fit’ shows a negative gradient

Part A: The most likely coefficient is -0.98. 
If the coefficient is -1, then each point would be exactly on the straight line (which they are not as shown on the graph). The graph however still shows a strong negative coefficient. It can be seen from the close distance of each point from the line of best fit. So -0.5 and -0.02 is unlikely as they show weak negative correlation

Part B: Refer to the second picture to see the horizontal and vertical distance between day 15 and day 20

The horizontal distance is 5 units
The vertical distance is read between 61 and 71.5, hence it’s 10.5
The slope = Vertical distance ÷ Horizontal Distance = 10.5÷5 = 2.1
The ‘downhill’ slope shows a negative gradient, hence the value of the slope is -2.1
The value of the slope shows that the surface area of the lake shrinks by 2.1 for every one day

Part C: The data in the table represents the relation between two variables. Since one variable doesn’t cause the change in the other variable, the data table represents correlation rather than a causation. 

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