Find the probability that in 200 tosses of a fair die, we will obtain at most 30 fives

Answers

Answer 1

Answer:

0.2946

Step-by-step explanation:

Number of tosses, n = 200

P(obtaining a 5), p = 1/6

q = 1 - p = 5/6

Normal approximation for binomial distribution

P(X < A) = P(Z < (A - mean)/standard deviation)

Mean = np

= 200 x 1/6

= 33.33

Standard deviation = √npq

= √(200(1/6)(5/6) )

= 5.27

P(at most 30 fives) = P(X ≤ 30)

= P(Z < (30.5 - 33.33)/5.27) (continuity correction of 0.5 is added to 30)

= P(Z < -0.54)

= 0.2946


Related Questions

please help me explain your answer only answer if you are sure

Answers

Answer:

The answer of top prism is 262

and down prism is 478

The upper figure is triangular prism.

so, we use bh+2ls+lb formula

B=5

h=3

s=4

l=19

Now,

surface area of triangular prism = bh+2ls+lb

= 5×3+2×19×4+19×5

= 262

The down figure is rectangular prism.

so, we use 2lw+2lh+2hw

l=5

h=6

w=19

Now,

The area of rectangular prism = 2lw+2lh+2hw

= 2×5×19+2×19×6+2×5×6

= 478

what is 2043.666666 rounded to 2 decimal places

Answers

Answer:

[tex]2043.67[/tex]

Step-by-step explanation:

Hundredths is at 2 decimal places.

The thousandths place is higher than 5, so add 1 to the hundredths place.

Answer:

2043.67

Step-by-step explanation:

If you’ve ever rounded a number, you would know that if it’s 5 or higher, round it up, and if it’s 4 or lower, round it down. In this case, the second decimal place reads ’6’ which is higher that 5, so we round up. The rest of the numbers stay the same

2043.67

HELP ME QUICK!! The best answer I will mark brainlest!

Answers

Answer: 1. (4, 8); 2. (3, 4)

Step-by-step explanation: I tried to get this to you fast but I can give you an explaination if you would like one :)

Suppose that y = 5 x plus 4 and it is required that y be within 0.005 units of 8. For what values of x will this be​ true?

Answers

Answer:

so we have an inequality for y -

7.995<y<8.005

Then now we need in inequality for x

(y-4)/5 = x

so that means that so if we have (7.995-4)/5 we get 3.995/5 = 7.99

so we have our first 7.99<x<b

Now we solve for b

So that means that 5.005/5 = 1.001

since we are changing it we switch our signs

from 7.99<x<1.001

we do 7.99>x>1.001

therefore

1.001<x<7.99

Answer:

0.795 [tex]\leq[/tex] y [tex]\leq[/tex] 0.805

Step-by-step explanation:

8 = 5x + 4

5x = 4

x = 4/5 or 0.800

therefore 0.800 + .005 and 0.800 - .005 =

0.795 [tex]\leq[/tex] y [tex]\leq[/tex] 0.805

A delivery truck is transporting boxes of two sizes: large and small. The large boxes weigh 50 pounds each, and the small boxes weigh 35 pounds each. There are 115 boxes in all. If the truck is carrying a total of 5000 pounds in boxes, how many of each type of box is it carrying?

Answers

Answer:

Step-by-step explanation:

First, "boxes of two sizes" means we can assign variables:

  Let x = number of large boxes

       y = number of small boxes

 

 "There are 115 boxes in all"    means    x + y = 115      [eq1]

 

Now, the pounds for each kind of box is:

    (pounds per box)*(number of boxes)

 

So,

  pounds for large boxes     +    pounds for small boxes      =    4125 pounds

                 "the truck is carrying a total of 4125 pounds in boxes"

 

        (50)*(x)                   +              (25)*(y)                  = 4125    [eq2]

 

It is important to find two equations so we can solve for two variables.

 

Solve for one of the variables in eq1 then replace (substitute) the expression for that variable in eq2.  Let's solve for x:

   x = 115 - y          [from eq1]

 

    50(115-y) + 25y = 4125             [from eq2]

     5750 - 50y  + 25y = 4125           [distribute]

     5750 - 25y = 4125

      -25y = -1625

         y = 65             [divide both sides by (-25)]

 

 There are 65 small boxes.

 

Put that value into either equation (now, which is easier?) to solve for x:

  x = 115 - y

  x = 115 - 65

  x = 50

 

  There are 50 large boxes.

 

Check (very important):

 

Is   50+65 = 115   ?       [eq1]

         115 = 115   ?yes

Is  50(50) + 25(65) = 4125   ?

     2500 + 1625 = 4125   ?

       4125 = 4125   ? ye

What’s the correct answer for this question?

Answers

Answer:

A: 97π/18 m

Step-by-step explanation:

Central Angle = 97°

In radians:

97° = 97π/180

Now

S = r∅

S = (10)(97π/180)

S = 97π/18 m

Answer:

The answer is option 1.

Step-by-step explanation:

Given that the formula for length of Arc is Arc = θ/360×2×π×r when r represents the radius of circle. Then, you have to substitute the following values into the formula :

[tex]arc = \frac{θ}{360} \times 2 \times \pi \times r[/tex]

Let θ = ∠VCW = 97°,

Let r = 10m,

[tex]arc = \frac{97}{360} \times 2 \times \pi \times 10[/tex]

[tex]arc = \frac{97}{360} \times 20 \times \pi[/tex]

[tex]arc = \frac{97}{18} \pi \: m[/tex]

Find the population mean or sample mean as indicated.
Sample: 17, 11, 8, 12, 22

Answers

Answer:

mean:12

Step-by-step explanation:

The population mean or sample mean as indicated in the given samples is 14

What is mean?

A mean in math is the average of a data set, found by adding all numbers together and then dividing the sum of the numbers by the number of numbers.

Mathematically,

Mean = Sum of the observations/number of observations

Now the given sample is,

17, 11, 8, 12, 22

So, Number of sample = 5

Thus, Mean = Sum of the sample /number of sample

Mean = (17 + 11 + 8 + 12 + 22) / 5

⇒ Mean = 70/5

⇒ Mean = 14

Thus, the population mean or sample mean as indicated in the given samples is 14

To learn more about mean :

https://brainly.com/question/21479395

#SPJ2

Use the table of values to find the line of regression and if justified at the 0.05 significance level, use it to find the predicted quality score of a TV set with a price of $1900. If the data does not suggest linear correlation, then use the average quality score as a prediction.
Price: 2,300, 1,800, 2,500, 2,700, 2,000, 1,700, 1,500, 2,700
Quality Score: 74, 73, 70, 66, 63, 62, 52, 68

Answers

Answer:

Step-by-step explanation:

no          x                y                   xy                               x²

1           2300         74             170200                      5290000

2          1800          73             131400                      3240000

3          2500         70             175000                     6250000

4          2700         66             178200                     7290000

5          2000         63             126000                     4000000

6          1700          62             105400                     2890000

7           1500         52              78000                      2250000

8           2700        68              183600                    7290000

Total   17200         528           1147800                   38500000

Mean of x is

[tex]\bar x = \frac{17200}{8} =2150[/tex]

Mean of y is

[tex]\bar y = \frac{528}{8} =66[/tex]

From the table above

we find [tex]\hat B_1[/tex]

[tex]\hat B_1=\frac{\sum xy- \bar x \sum y}{\sum x^2- n \barx^2} \\\\=\frac{1147800-2150(528)}{38500000-8(2150)^2} \\\\=\frac{1147800-1135200}{38500000-36980000} \\\\=\frac{12600}{1520000} \\\\=0.008289[/tex]

so [tex]\hat b_0[/tex] is

[tex]\hat b_0=\bar y-\bar B_1 x\\\\=66-0.008289(2150)\\\\=66-17.82135\\\\=48.17865[/tex]

The line of regression is

The price x is 1900

[tex]\hat y =\hat B_0+\hat B_1x\\\\=48.17865+0.008289\times1900\\\\=48.17865+15.7491\\\\=63.928[/tex]

The line of regression is 63.928

Answer:

y=48.2+0.00829x;64

Step-by-step explanation:

Tameka can make 32 beads with 4 sheets of paper. Right now she only has 3 sheets of paper. How many beads can Tameka make?

Answers

Answer:

24 beads

Step-by-step explanation:

beads in 4 sheet=32

beads in 1 sheet=32/4=8

beads in 3 sheet=8*3=24

Answer:

Assuming that you can make the same number of beads with 1 sheet of paper, Tameka can make 24 beads with 3 sheets of paper.

For which x is f(x)?=-3
-7
-4
4
5

Answers

Answer:

B.

✔ -4

Step-by-step explanation:

E 2021



Help Me PLEASE!!!


A card is chosen at random from a standard deck of 52 cards, and then it is replaced and another card is chosen. What is the probability that at least one of the cards is a diamond or an ace?

Answers

Answer:

P = 0.5207

Step-by-step explanation:

First, we have three options: Just the first card is a diamond or an ace, Just the second card is a diamond or an ace and both cards are diamonds or aces.

Additionally, there are 16 cards that are diamond or aces in a standard deck of 52 cards (13 diamonds and 3 aces that are not diamonds). It means that there are 36 cards that are not diamond or aces (52 - 16 = 36).

So, the probability that just the first card is a diamond or an ace is calculated as:

[tex]P_1=\frac{16}{52}*\frac{36}{52}=0.2130[/tex]

At the same way, the probability that just the second card is a diamond or an ace is:

[tex]P_2=\frac{36}{52}*\frac{16}{52}=0.2130[/tex]

Finally, the probability that both cards are diamonds or aces is:

[tex]P_3=\frac{16}{52}*\frac{16}{52}=0.0947[/tex]

Therefore, the probability that at least one of the cards is a diamond or an ace is:

[tex]P=P_1+P_2+P_3\\P=0.2130+0.2130+0.0947\\P=0.5207[/tex]

The percent, X , of shrinkage o n drying for a certain type of plastic clay has an average shrinkage percentage :, where parameter : is unknown. A random sample of 45 specimens from this clay showed an average shrinking percentage of 18.4 and a standard deviation of 2.2.

Required:
a. Estimate at 5% level of significance whether the true average shrinkage percentage U: is greater than 17.5 and write your conclusion.
b. Report the p-value.

Answers

Answer:

a) [tex]t=\frac{18.4-17.5}{\frac{2.2}{\sqrt{45}}}=2.744[/tex]    

The degrees of freedom are given by:

[tex]df=n-1=45-1=44[/tex]  

The critical value for this case is [tex]t_{\alpha}=1.68[/tex] since the calculated value is higher than the critical we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly higher than 18.4

b) [tex]p_v =P(t_{(44)}>2.744)=0.0044[/tex]  

Step-by-step explanation:

Information given

[tex]\bar X=18.4[/tex] represent the sample mean

[tex]s=2.2[/tex] represent the sample standard deviation

[tex]n=45[/tex] sample size  

[tex]\mu_o =17.5[/tex] represent the value to verify

[tex]\alpha=0.05[/tex] represent the significance level

t would represent the statistic

[tex]p_v[/tex] represent the p value

Part a

We want to test if the true mean is higher than 17.5, the system of hypothesis would be:  

Null hypothesis:[tex]\mu \leq 17.5[/tex]  

Alternative hypothesis:[tex]\mu > 17.5[/tex]  

The statistic is given by:

[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex]  (1)  

And replacing we got:

[tex]t=\frac{18.4-17.5}{\frac{2.2}{\sqrt{45}}}=2.744[/tex]    

The degrees of freedom are given by:

[tex]df=n-1=45-1=44[/tex]  

The critical value for this case is [tex]t_{\alpha}=1.68[/tex] since the calculated value is higher than the critical we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly higher than 18.4

Part b

The p value would be given by:

[tex]p_v =P(t_{(44)}>2.744)=0.0044[/tex]  

In a survey of students, 60% were in high school and 40% were in middle school. Of the high school students, 30% had visited a foreign country. If a surveyed student is selected at random, what is the probability that the student is in high school and has visited a foreign country?

Answers

Answer:

The probability that the student is in high school and has visited a foreign country is 0.18.

Step-by-step explanation:

We are given that in a survey of students, 60% were in high school and 40% were in middle school.

Of the high school students, 30% had visited a foreign country.

Let the Probability that students were in high school = P(H) = 60%

Probability that students were in middle school = P(M) = 40%

Also, let F = event that students had had visited a foreign country

So, Probability that high school students had visited a foreign country = P(F/H) = 30%

Now, probability that the student is in high school and has visited a foreign country is given by = Probability that students were in high school [tex]\times[/tex] Probability that high school students had visited a foreign country  

            =  P(H)  [tex]\times[/tex]  P(F/H)

            =  0.60  [tex]\times[/tex]  0.30

            =  0.18 or 18%

Write a quadratic function f whose zeros are −6 and −1.

Answers

F(X)=(X+6)(X+1)=0 That’s all!!!

Answer:

y = (x+6) (x+1) or in quadratic form: y = x² + 7x + 6

Step-by-step explanation:

New York City is known for it's tourist attractions and high priced real estate. The mean hotel room rate is $202 per night. Assume that the room rates are normally distributed with a standard deviation of $70.What is the probability that a hotel room costs between $210 and $290?

Answers

Answer:

[tex]P(210<X<290)=P(\frac{210-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{290-\mu}{\sigma})=P(\frac{210-202}{70}<Z<\frac{290-202}{70})=P(0.114<z<1.26)[/tex]

And we can find this probability with this difference

[tex]P(0.114<z<1.26)=P(z<1.26)-P(z<0.114)[/tex]

And we can find the difference with the normal standard distirbution or excel:

[tex]P(0.114<z<1.26)=P(z<1.26)-P(z<0.114)=0.896-0.545=0.351[/tex]

Step-by-step explanation:

Let X the random variable that represent the hotel room cost of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(202,70)[/tex]  

Where [tex]\mu=202[/tex] and [tex]\sigma=70[/tex]

We are interested on this probability

[tex]P(210<X<290)[/tex]

The z score formula is given by:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

Using the formula we got:

[tex]P(210<X<290)=P(\frac{210-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{290-\mu}{\sigma})=P(\frac{210-202}{70}<Z<\frac{290-202}{70})=P(0.114<z<1.26)[/tex]

And we can find this probability with this difference

[tex]P(0.114<z<1.26)=P(z<1.26)-P(z<0.114)[/tex]

And we can find the difference with the normal standard distirbution or excel:

[tex]P(0.114<z<1.26)=P(z<1.26)-P(z<0.114)=0.896-0.545=0.351[/tex]

4 ounces for every 16 ounces. Rate or ratio and in simplest form

Answers

4 for 16 would be written as 4/16.

Dividing both numbers by 4 it is simplified to 1/4

4 ounces for every 16 ounces can be written as:

[tex]\displaystyle \frac{4}{16} =\frac{1}{4}[/tex]

As a ratio, it can be expressed as:

[tex]1:4[/tex]

6(4x - 3) - 30
24x - 18 = 30
24% -18 + 18 = 30 + 18
24x = 48
24x 48
24 24
X = 2
Original Equation
Step 1
Step 2
Step 3
Step 4
Step 5
Which of these is not part of the solution process?
A. Using the associative property
B. Adding 18 to both sides to isolate the variable term
C. Dividing both sides by 24 to isolate the variable
D. Using the distributive property

Answers

Answer:

A-using the associative property

Step-by-step explanation:



Daniel deposits $300 into an account that earns 16% interest annually. Which equation can be used to model his account balance, y, after x years?

Answers

Answer:

[tex]y=300(1+0.16)^x[/tex]

Step-by-step explanation:

This account  can be  modeled using the compound interest formula.the compound interest formula is expressed as

 [tex]A= P(1+r )^t[/tex]

Where

A =final amount  = y

P=initial principal balance = $300

r=interest rate  = 16%= 0.16

t=number of time periods elapsed= x

Hence the equation to model his account balance/ final amount A (y) after time (x) years is

[tex]y=300(1+0.16)^x[/tex]

76,80,88,95,100,101,? Which number comes next in this sequence?

Answers

Answer:

112

Step-by-step explanation:

Difference between each 4,8,7,5,1

Add numbers next to each other in pairs = 12

So 12-1= 11 and

101+11=112

What’s the correct answer for this?

Answers

Answer:

D.

Step-by-step explanation:

Since opposite angles of a quadrilateral inscribes in a circle add up to 180°

So,

<P + <N = 180°

2x+2x-12 = 180°

4x = 180+12

4x = 192

Dividing both sides by 4

x = 48

Now

<P = 2(48)

<P = 96

Now

<N = 2(48)-12

<N = 96-12

<N = 84

Plz help me :(

Prove for any value of y the value of expression y^4−(y^2−9)(y^2+9) is divisible by 9.

Answers

PROOF:

[tex]y^4-(y^2-9)(y^2+9)\\=y^4 -(y^4-81)\\=y^4-y^4+81\\=81[/tex]

And 81 is ALWAYS divisible by 9

Answer:

Its 9*9

Step-by-step explanation:

The equation equals 81 and 9*9=81.

Have a great day! :)

An individual closes out help desk tickets at a rate of 4 tickets per hour for h hours. Write an equation that expresses the situation, let x be the independent variable and y be the dependent variable

Answers

Answer:

y=4x

Step-by-step explanation:

an independent variable is the variable  that is changed or controlled in an experiment or observation to test the effects on the dependent variable

a dependent variable is variable being tested and measured in a scientific experiment

in this case, the number of help desk tickets  closed out is dependent on the number of hours the individual works so y is the number of tickets closed (dependent variable). The number of tickets closed of will be 4 multiplied by the number of hours worked i.e. y=4x

22,056 people went to the baseball game on Sunday. Half as many people came on money. How many people were at the baseball game on Sunday and Monday altogether?

Answers

Answer:

33084

Step-by-step explanation:

22056 divided by 2 =11028

altogether (on sunday and monday) the total amount would be..

22056+11028=33084

Answer:

33084

Step-by-step explanation:

If 22056 people came to the game on Sunday and Half as many people came on Monday, you do

22056 divided by 2. this is how many people cam on monday

Add this answer to 22056 and this is how many people came on both days.

The present value of a perpetuity paying 1 every two years with first payment due immediately is 7.21 at an annual effective rate of i. Another perpetuity paying R every three years with the first payment due at the beginning of year two has the same present value at an annual effective rate of i + 0.01.
Calculate R.
(A) 1.23
(B) 1.56
(C) 1.60
(D) 1.74
(E) 1.94

Answers

Answer:

Step-by-step explanation:

image attached (representing first perpetuity on number line)

Present value is 7.21

[tex]7.21=\frac{1}{1-u^2} \\\\1-\frac{1}{7.21} =u^2\\\\\frac{6.21}{7.21} =(1+i)^{-2}\\\\(1+i)^2=\frac{7.21}{6.21} \\\\(i+1)=\sqrt{\frac{7.21}{6.21} }\\\\ i=\sqrt{\frac{7.21}{6.21} } -1\\\\=0.77511297[/tex]

image attached (representing second perpetuity on number line)

we have ,

[tex]7.21=\frac{Ru}{1-u^3}[/tex]

Here,

[tex]V=\frac{1}{1+i}[/tex]i

i = 0.077511297 + 0.01

[tex]\therefore V =\frac{1}{1.087511295} =(1.087511297)^-^1\\\\7.21=\frac{R(1.087511297)^-^1}{1-(1.087511297)^-^3} \\\\7.21=4.132664645R\\\\R=\frac{7.21}{4.132664645} \\\\R= 1.7446370\approx1.74[/tex]

Therefore, value of R is 1.74


Determine 6m 9m how much greater the area of the yellow rectangle is than the area of the gree rectangle 2m 5m

Answers

Step-by-step explanation:

multiple 6 by 9 then 2 by 5 then subtract them

Answer:

44

Step-by-step explanation:

(6*9) - (2*5)

54 - 10

44

A gumball machine has 100 red gumballs. If the red gumballs are 25% of the total number of gumballs, how many gumballs are in the gumball machine?

Answers

Answer: 400

Step-by-step explanation:

25% is equal to one quarter (1/4). If theres 100 red gumballs then there must be 300 more gumballs in the machine because a quarter of a number is always even.

How do u solve this?

Answers

Answer:

0

Step-by-step explanation:

Tuesday : -1/2

Wednesday + 3/4

Thursday : -3/8

Add them together

-1/2 + 3/4- 3/8

Get a common denominator

-4/8 + 6/8 - 3/8

-1/8

The closest integer value to -1/8 is 0

What is a word problem for 15 minus 28?

Answers

Answer:

A word problem for that would be Sam had 28 chocolates and Bob took away 15. How many does Sam have left?

Step-by-step explanation:

I don't know how to show work for writing a word problem. Sorry

Answer:

Step-by-step explanation:

Jane has $15 in her bank account. She wrote a $28 check for buying a fiction book. How much is her balance now?

Temperature transducers of certain type are shipped in batches of 50. A sample of 60 batches was selected, and the number of transducers in each batch not conforming to design specifications was determined, resulting in the following data:

2 1 2 3 1 1 3 2 0 5 3 3 1 3 2 4 7 0 2 3
0 4 2 1 3 1 1 3 41 2 3 2 2 8 4 5 1 3 1
5 0 2 3 2 1 0 6 4 2 1 6 0 3 3 3 6 2 3

a. Determine frequencies and relative frequencies for the observed values of x = number of non-conforming transducers in a batch. (Round your relative frequencies to three decimal places.)
b. What proportion of batches in the sample have at most four non-conforming transducers? (Round your answer to three decimal places.)

Answers

Answer:

a.

Number:      0,    1,       2,     3,      4,       5,    6,    7,     8

Frequency:  6,    12,     13,     15,     5,      3,     3,    1,    1

b. The proportion of the batches that have at most is 0.864

Step-by-step explanation:

a. The given data are;

2 1 2 3 1 1 3 2 0 5 3 3 1 3 2 4 7 0 2 3

0 4 2 1 3 1 1 3 4 1 2 3 2 2 8 4 5 1 3 1

5 0 2 3 2 1 0 6 4 2 1 6 0 3 3 3 6 2 3

The frequencies are;

x     fx  

0     6

1      12

2     13

3      15

4      5

5      3

6      3

7       1

8       1

The relative frequency are;

x     Rfx  

0   0.102

1   0.203

2   0.220

3   0.254

4   0.085

5   0.051

6   0.051

7   0.017

8   0.017

b. The proportion of the batches that have at most 4 is given as follows;

The number of the batches that have at most 4 = 6 + 12 + 13 + 15 + 5 = 51

Therefore, the proportion of the batches that have at most 4 = 51 / 59 = 0.864.

Consider the following set of sample data.
18 26 30 42 50 52 52 76 78 84
For the given data, the mean is_______, the median is________, and the mode is_______.
Suppose the value 76 in the data is mistakenly recorded as 55 instead of 76. For the sample with this error, the mean is_________, the median is______, and the mode is_______. The mean_____, the median_______, and the mode______. Suppose the value 76 in the original sample is inadvertently removed from the sample. For the sample with this value removed, the mean is_______, the median is_______, and the mode is________. The mean_________, the median_______, and the mode________.

Answers

Answer:

For the given data, the mean is 50.8, the median is 51, and the mode is 52.

For the sample with this error, the mean is 48.7, the median is 51, and the mode is 52.

For the sample with this value removed, the mean is 43.2, the median is 50, and the mode is 52.

Step-by-step explanation:

We are given the following set of sample data below;

18, 26, 30, 42, 50, 52, 52, 76, 78, 84.

The formula for calculating mean is given by;

         Mean  =  [tex]\frac{\text{Sum of all data values}}{\text{Total number of observations}}[/tex]

                     =  [tex]\frac{18+ 26+ 30+ 42+ 50+ 52+ 52+ 76+ 78+ 84}{10}[/tex]  

                     =  [tex]\frac{508}{10}[/tex]  =  50.8

For calculating median, we have to observe that the number of observations (n) in our data is even or odd, i.e;

If n is odd, then the formula for calculating median is given by;

                    Median  =  [tex](\frac{n+1}{2})^{th} \text{ obs.}[/tex]

If n is even, then the formula for calculating median is given by;

                    Median  =  [tex]\frac{(\frac{n}{2})^{th}\text{ obs.} +(\frac{n}{2}+1)^{th}\text{ obs.} }{2}[/tex]

Now, here in our data the number of observations is even, i.e. n = 10.

So, Median  =  [tex]\frac{(\frac{n}{2})^{th}\text{ obs.} +(\frac{n}{2}+1)^{th}\text{ obs.} }{2}[/tex]

                    =  [tex]\frac{(\frac{10}{2})^{th}\text{ obs.} +(\frac{10}{2}+1)^{th}\text{ obs.} }{2}[/tex]

                    =  [tex]\frac{5^{th}\text{ obs.} +6^{th}\text{ obs.} }{2}[/tex]

                    =  [tex]\frac{50+52 }{2}[/tex]  =  51

A Mode is a value that appears the maximum number of times in our data.

In our data, the value 52 is appera]ing maximum number of times, i.e. 2 times which means that mode of our data is 52.

Now, suppose the value 76 in the data is mistakenly recorded as 55 instead of 76. For the sample with this error,

Mean will be changed as value has been changed.

            New Mean  =   [tex]\frac{18+ 26+ 30+ 42+ 50+ 52+ 52+ 55+ 78+ 84}{10}[/tex]  

                                =  [tex]\frac{487}{10}[/tex]  =  48.7

There will be no change in median because there is no change in the 5th and 6th observation of the data.

Also, there will be no change in mode as stiil 52 appears maximum number of times in our data.

Now, suppose the value 76 in the original sample is inadvertently removed from the sample. For the sample with this value removed,

Mean will be changed as value has been removed from data.

            New Mean  =   [tex]\frac{18+ 26+ 30+ 42+ 50+ 52+ 52+ 78+ 84}{9}[/tex]  

                                =  [tex]\frac{432}{10}[/tex]  =  43.2

Median will also get changed because the number of observation is now odd, i.e. n = 9

            So, Median  =  [tex](\frac{n+1}{2})^{th} \text{ obs.}[/tex]

                                 =  [tex](\frac{9+1}{2})^{th} \text{ obs.}[/tex]

                                 =  [tex]5^{th} \text{ obs.}[/tex] = 50

Also, there will be no change in mode as stiil 52 appears maximum number of times in our data.
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