What is the outlier for the data set? 19, 19, 27, 21, 77, 18, 23, 29

Three populations have proportions 0.1, 0.3, and 0.5. We select random samples of the size n from these populations. Only two of

IRINA_888 [86]

Answer:

(1) A Normal approximation to binomial can be applied for population 1, if n = 100.

(2) A Normal approximation to binomial can be applied for population 2, if n = 100, 50 and 40.

(3) A Normal approximation to binomial can be applied for population 2, if n = 100, 50, 40 and 20.

Step-by-step explanation:

Consider a random variable X following a Binomial distribution with parameters n and p.

If the sample selected is too large and the probability of success is close to 0.50 a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:

np ≥ 10
n(1 - p) ≥ 10

The three populations has the following proportions:

p₁ = 0.10

p₂ = 0.30

p₃ = 0.50

(1)

Check the Normal approximation conditions for population 1, for all the provided n as follows:

$n_{a}p_{1}=10\times 0.10=1$

Thus, a Normal approximation to binomial can be applied for population 1, if n = 100.

(2)

Check the Normal approximation conditions for population 2, for all the provided n as follows:

$n_{a}p_{1}=10\times 0.30=310\\\\n_{c}p_{1}=50\times 0.30=15>10\\\\n_{d}p_{1}=40\times 0.10=12>10\\\\n_{e}p_{1}=20\times 0.10=6$

Thus, a Normal approximation to binomial can be applied for population 2, if n = 100, 50 and 40.

(3)

Check the Normal approximation conditions for population 3, for all the provided n as follows:

$n_{a}p_{1}=10\times 0.50=510\\\\n_{c}p_{1}=50\times 0.50=25>10\\\\n_{d}p_{1}=40\times 0.50=20>10\\\\n_{e}p_{1}=20\times 0.10=10=10$

Thus, a Normal approximation to binomial can be applied for population 2, if n = 100, 50, 40 and 20.

8 0

3 years ago

Please help me out!!!!

LuckyWell [14K]

Answer:

the value for x is

$11 \sqrt{3}$

and the value for y is 11 on the triangle problem we use sin60 to find x and we use tan60 to find y

7 0

3 years ago

A school is preparing a trip for 400 students. The company who is providing the transportation has 10 buses of 50 seats each and

Naya [18.7K]

<h2>The number of small buses used = 5</h2><h2>The number of big buses used = 4</h2>

Step-by-step explanation:

Let us assume the total number of small buses needed = x

The capacity of 1 small bus = 40

So, the capacity of x buses = 40(x) = 40 x

Let us assume the total number of big buses needed = y

The capacity of 1 big bus = 50

So, the capacity of y buses = 50(y) = 50 y

Also, the total students travelling = 400

So, the number of students traveling by (Small bus + Big bus) = 400

⇒ 40 x + 50 y = 400 ..... (1)

Also, the total number of drivers available = 9

⇒ x + y = 9 ..... (2)

Also, x ≤ 8, y ≤ 10

Now, solving both equations, we get:

40 x + 50 y = 400 ..... (1)

x + y = 9 ⇒ y = (9-x) put in (1)

40 x + 50 y = 400 ⇒ 40 x + 50 (9-x) = 400

or, 40 x + 450 - 50 x = 400

or, - 10 x =- 50

or, x = 5 ⇒ y = (9-x) = 9- 5 = 4

Hence the number of small buses used = 5

The number of big buses used = 4

8 0

3 years ago

Fr33 points 38/827. 72/727

Virty [35]

Answer:

thx

Step-by-step explanation:

:D:D:D

7 0

3 years ago

Read 2 more answers

A radio station advertises a contest with ten cash prizes totaling $5510. There is to be a $100 difference between each successi

My name is Ann [436]

Answer:

option C

c. least: $101

greatest: $1001

Step-by-step explanation:

A radio station advertises a contest with ten cash prizes totaling $5510. There is to be a $100 difference between each successive prize.

Sum of 10 prizes = 5510

100 is the difference. there is a common difference d=100

So its a arithmetic sequence

the sum formula for arithmetic sequence is

$S_n = \frac{n}{2}(2a_1 +(n-1)d)$

Sn = 5510, n=10 n d= 100 we need to find out first term a1

$5510 = \frac{10}{2}(2a_1 +(10-1)100)$

5510 = 5 (2a1 + 900)

5510 = 10a1 + 4500

Subtract 4500 on both sides

1010= 10a1

divide by 10 on both sides

a1 = 101

so first term that is least term is 101

To find out greatest term we use formula

$a_n = a_1 + (n-1) d$

a(10) = 101 + (10-1)100

= 101 + 900= 1001

greatest is 1001

4 0

4 years ago