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Lostsunrise [7]
2 years ago
14

A group of friends has gotten very competitive with their board game nights. They have found that overall, they each have won an

average of 18 games, with a population standard deviation of 6 games. If a sample of only 2 friends is selected at random from the group, select the expected mean and the standard deviation of the sampling distribution from the options below. Remember to round to the nearest whole number.
Mathematics
1 answer:
Amanda [17]2 years ago
7 0

Answer: \mu_x=18\text{ hours}

\sigma_x=4\text{ hours}

Step-by-step explanation:

We know that mean and standard deviation of sampling distribution is given by :-

\mu_x=\mu

\sigma_x=\dfrac{\sigma}{\sqrt{n}}

, where \mu = population mean

\sigma =Population standard deviation.

n= sample size .

In the given situation, we have

\mu=18\text{ hours}

\sigma=6\text{ hours}

n= 2

Then, the expected mean and the standard deviation of the sampling distribution will be :_

\mu_x=\mu=18\text{ hours}

\sigma_x=\dfrac{\sigma}{\sqrt{n}}=\dfrac{6}{\sqrt{2}}=4.24264068712\approx4  [Rounded to the nearest whole number]

Hence, the the expected mean and the standard deviation of the sampling distribution :

\mu_x=18\text{ hours}

\sigma_x=4\text{ hours}

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There are 8 sophomores on the academic team. At the last competition, they each took the math test. Their scores were 82%, 92%,
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3 0
2 years ago
among a group of students 50 played cricket 50 played hockey and 40 played volleyball. 15 played both cricket and hockey 20 play
kondaur [170]

Answer:

Cricket only= 30

Volleyball only = 15

Hockey only = 25

Explanation:

Number of students that play cricket= n(C)

Number of students that play hockey= n(H)

Number of students that play volleyball = n(V)

From the question, we have that;

n(C) = 50, n(H) = 50, n(V) = 40

Number of students that play cricket and hockey= n(C∩H)

Number of students that play hockey and volleyball= n(H∩V)

Number of students that play cricket and volleyball = n(C∩V)

Number of students that play all three games= n(C∩H∩V)

From the question; we have,

n(C∩H) = 15

n(H∩V) = 20

n(C∩V) = 15

n(C∩H∩V) = 10

Therefore, number of students that play at least one game

n(CᴜHᴜV) = n(C) + n(H) + n(V) – n(C∩H) – n(H∩V) – n(C∩V) + n(C∩H∩V)

= 50 + 50 + 40 – 15 – 20 – 15 + 10

Thus, total number of students n(U)= 100.

Note;n(U)= the universal set

Let a = number of people who played cricket and volleyball only.

Let b = number of people who played cricket and hockey only.

Let c = number of people who played hockey and volleyball only.

Let d = number of people who played all three games.

This implies that,

d = n (CnHnV) = 10

n(CnV) = a + d = 15

n(CnH) = b + d = 15

n(HnV) = c + d = 20

Hence,

a = 15 – 10 = 5

b = 15 – 10 = 5

c = 20 – 10 = 10

Therefore;

For number of students that play cricket only;

n(C) – [a + b + d] = 50 – (5 + 5 + 10) = 30

For number of students that play hockey only

n(H) – [b + c + d] = 50 – ( 5 + 10 + 10) = 25

For number of students that play volleyball only

n(V) – [a + c + d] = 40 – (10 + 5 + 10) = 15

3 0
2 years ago
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