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Oksana_A [137]
3 years ago
6

In a class of P students, the average of test scores is 70. In another class the average test scores is 92. When scores of the t

wo classes are combined the average of test scores is 86. what is the value of p/n
Mathematics
1 answer:
Basile [38]3 years ago
7 0

So, since averages are defined as:

\frac{\sum_{k=1}^{P} P_k}{P}=70

So, since P are the total number of elements and P_k is the P_kth student. This is saying if we sum over each student's score and divide it by the number of students, we should get P, which is true.

So, using that logic, the other class can be represented as:

\frac{\sum_{k=1}^{N} N_k}{N}=70

We can take both of these equations and multiply them by N:

\sum_{k=1}^{P} P_k=70P

\sum_{k=1}^N N_k=92N

So, if we want to find the average of this we should add both our equations then divide by P+N, which is the number of all the students.

\frac{\sum_{k=1}^{P} P_k+\sum_{k=1}^{N}N_k}{N+P}=\frac{70P+92N}{N+P}

To make this simpler we can replace our LHS with 86, since that's the average of both classes combined.

86=\frac{92N+70P}{N+P} \implies\\ 86N+86P=92N+70P \implies \\ 16P=6N \implies \\ \frac{16P}{N}=6 \implies \\ \frac{P}{N}=\frac{6}{16}

Simplified we would have P/N=3/8.

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A programmer plans to develop a new software system. In planning for the operating system that he will​ use, he needs to estimat
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Answer:

a) n = 9604

b) n = 381

Step-by-step explanation:

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\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

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The margin of error is:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

90% confidence level

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a. Assume that nothing is known about the percentage of computers with new operating systems. n = ?

When we do not know the proportion, we use \pi = 0.5, which is when we are going to need the largest sample size.

The sample size is n when M = 0.01.

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.01 = 1.96\sqrt{\frac{0.5*0.5}{n}}

0.01\sqrt{n} = 1.96*0.5

\sqrt{n} = \frac{1.96*0.5}{0.01}

(\sqrt{n})^{2} = (\frac{1.96*0.5}{0.01})^{2}

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b. Assume that a recent survey suggests that 99% of computers use a new operating system. n = ?

Now we have that \pi = 0.99. So

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0.01\sqrt{n} = 1.96*\sqrt{0.99*0.01}

\sqrt{n} = \frac{1.96*\sqrt{0.99*0.01}}{0.01}

(\sqrt{n})^{2} = (\frac{1.96*\sqrt{0.99*0.01}}{0.01})^{2}

n = 380.3

Rouding up

n = 381

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