All categories

3 years ago

Calculate the variance for the following population of N = 5 scores: 2, 13, 4, 10, 6

Mathematics

1 answer:

AnnyKZ [126]3 years ago

4 0

Answer:

Step-by-step explanation:

(1) Get the mean: (2 + 13 + 4 + 10 + 6) / 5 = 7

(2) Variance Formula: Σ(x-mean)^2)/N

-> <u>(2-7)^2 + (13-7)^2 + (4-7)^2 + (10-7) ^2 + (6-7)^2</u>

= 16

You might be interested in

You have a $3175 student loan. Suppose you pay the loan off in two and a half years (30 months). Estimate your monthly payment b

finlep [7]

Answer:

$106

Step-by-step explanation:

The formula given for Monthly payment of a loan =

P × [ r (1 + r)/(1 + r)^n - 1

Where

r = interest rate

n = number of monthly payments

P = Present value of the loan

From the question,

r = interest rate, we were told to ignore hence, r = 0

P = $3,175

n = 30

Hence,

Amount to be paid monthly = P/n

= $3175/30

= $105.83

Approximately to the nearest dollars

= $106

6 0

3 years ago

Godfrey plays a game in which he throws two fair six sided dice. If he rolls two sizes, he wins 20p, if he rolls one six, he win

blagie [28]

Answer:

The Probabilty distribution for the amount Godfrey gains in one turn is then given as

X ||| P(X)

15p | 0.0278

5p | 0.278

-5p | 0.6942

Step-by-step explanation:

If random variable X represents the amount Godfrey gains in one turn.

There are 3 different possible outcomes for X.

- Godfrey pays 5p to enter the game and gets two sixes and wins 20p.

Net gain = 15p

Probability of getting two sixes from two fair dice

= (number of outcomes with two sixes) ÷ (total number of outcomes)

number of outcomes with two sixes = 1

total number of possible outcomes = 36

Probability of getting two sides from two fair dice = (1/36) = 0.0278

- Godfrey pays 5p to enter the game and gets only one six and wins 10p.

Net gain = 5p

Probability of getting one six from either of two fair dice

= (number of outcomes with one six) ÷ (total number of outcomes)

number of outcomes with one six = 2 × n[(6,1), (6,2), (6,3), (6,4), (6,5)] = 2 × 5 = 10

total number of possible outcomes = 36

Probability of getting two sides from two fair dice = (10/36) = 0.278

- Godfrey pays 5p to enter the game and doesn't win anything

Net gain = -5p

Probability of not getting two sixes or one six.

= 1 - [(Probability of getting two sixes) + (Probability of getting one six on.wither dice)]

= 1 - 0.0278 - 0.278 = 0.6942

Probability of getting not getting two sixes or one six = 0.6942

The Probabilty distribution for the amount Godfrey gains in one turn is then given as

X ||| P(X)

15p | 0.0278

5p | 0.278

-5p | 0.6942

Hope this Helps!!!

4 0

3 years ago

Read 2 more answers

A swimmer is racing to the other side of the

kiruha [24]

Either B or C. They look exactly the same but I can’t tell because the photo is blurry

y intercept is 75 and x intercept is 75/2.5=30

4 0

3 years ago

How to solve 76×327 step by step

hram777 [196]

Here's a method that takes advantage of the values of these particular numbers.

... 76 = 80 - 4

... 76 × 327 = (80 -4)×327

... = 80×327 - 4×327

Repeated doubling will give us values that are 2, 4, and 8 times 327.

... 2×327 = 327+327 = 654

... 2×654 = 4×327 = 654+654 = 1308 . . . . we'll use this later

... 2×1308 = 8×327 = 2616

We want 80×327, so we can add a zero to the end of this last:

... 80×327 = 26160

Now, we can subtract 4×327 to get 76×327

... 80×327 - 4×327 = 26160 -1308 = 24852 = 76×327

_____

More conventionally, you would multiply every digit of one number by every digit of the other and add the products according to their respective place values.

327 × 076 = (3×0)×10000 + (3×7 +2×0)×1000 +(3×6 +7×0 +2×7)×100 +(2×6 +7×7)×10 +(7×6)×1

... = 0 +21,000 +3,200 + 610 +42

... = 24,852

Note the pattern of partial products here. This is a method taught to/by practitioners of Vedic mathematics, and can be done in your head. At most, you would write down the partial product sums 21, 32, 61, and 42 to keep from having to carry more than one number in your head at a time.

5 0

3 years ago

Lesson 12 practice problem number 2 imath book unit 1 geomitry

Nostrana [21]

Ur not telling us how to solve or what to solve.

6 0

3 years ago