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3 years ago

All students = 700

Mathematics

1 answer:

Sedaia [141]3 years ago

7 0

Answer:

See below

Step-by-step explanation:

Given:

Total number = 700
Off grid class = 450
Online class = 320
Both = 275

Based on the given we find the following

1) Off-grid class only:

450 - 275 = 175

3) Online class only:

320 - 275 = 45

2) None of the classes:

700 - (175 + 275 + 45) = 205

4) Not online:

700 - 320 = 380

5) Not off-grid:

700 - 450 = 250

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The Institute of Management Accountants (IMA) conducted a survey of senior finance professionals to gauge members’ thoughts on g

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Answer:

(1) The probability that the sample percentage indicating global warming is having a significant impact on the environment will be between 64% and 69% is 0.3674.

(2) The two population percentages that will contain the sample percentage with probability 90% are 0.57 and 0.73.

(3) The two population percentages that will contain the sample percentage with probability 95% are 0.55 and 0.75.

Step-by-step explanation:

Let X = number of senior professionals who thought that global warming is having a significant impact on the environment.

The random variable X follows a Binomial distribution with parameters n = 100 and p = 0.65.

But the sample selected is too large and the probability of success is close to 0.50.

So a Normal approximation to binomial can be applied to approximate the distribution of p if the following conditions are satisfied:

np ≥ 10
n(1 - p) ≥ 10

Check the conditions as follows:

$np= 100\times 0.65=65>10\\n(1-p)=100\times (1-0.65)=35>10$

Thus, a Normal approximation to binomial can be applied.

So, $\hat p\sim N(p, \frac{p(1-p)}{n})=N(0.65, 0.002275)$ .

(1)

Compute the value of $P(0.64$ as follows:

$P(0.64$

$=P(-0.20$

Thus, the probability that the sample percentage indicating global warming is having a significant impact on the environment will be between 64% and 69% is 0.3674.

(2)

Let $p_{1}$ and $p_{2}$ be the two population percentages that will contain the sample percentage with probability 90%.

That is,

$P(p_{1}$

Then,

$P(p_{1}$

$P(\frac{p_{1}-p}{\sqrt{\frac{p(1-p)}{n}}}$

$P(-z$

The value of z for P (Z < z) = 0.95 is

z = 1.65.

Compute the value of $p_{1}$ and $p_{2}$ as follows:

$-z=\frac{p_{1}-p}{\sqrt{\frac{p(1-p)}{n}}}\\-1.65=\frac{p_{1}-0.65}{\sqrt{\frac{0.65(1-0.65)}{100}}}\\p_{1}=0.65-(1.65\times 0.05)\\p_{1}=0.5675\\p_{1}\approx0.57$ $z=\frac{p_{2}-p}{\sqrt{\frac{p(1-p)}{n}}}\\1.65=\frac{p_{2}-0.65}{\sqrt{\frac{0.65(1-0.65)}{100}}}\\p_{2}=0.65+(1.65\times 0.05)\\p_{1}=0.7325\\p_{1}\approx0.73$

Thus, the two population percentages that will contain the sample percentage with probability 90% are 0.57 and 0.73.

(3)

Let $p_{1}$ and $p_{2}$ be the two population percentages that will contain the sample percentage with probability 95%.

That is,

$P(p_{1}$

Then,

$P(p_{1}$

$P(\frac{p_{1}-p}{\sqrt{\frac{p(1-p)}{n}}}$

$P(-z$

The value of z for P (Z < z) = 0.975 is

z = 1.96.

Compute the value of $p_{1}$ and $p_{2}$ as follows:

$-z=\frac{p_{1}-p}{\sqrt{\frac{p(1-p)}{n}}}\\-1.96=\frac{p_{1}-0.65}{\sqrt{\frac{0.65(1-0.65)}{100}}}\\p_{1}=0.65-(1.96\times 0.05)\\p_{1}=0.552\\p_{1}\approx0.55$ $z=\frac{p_{2}-p}{\sqrt{\frac{p(1-p)}{n}}}\\1.96=\frac{p_{2}-0.65}{\sqrt{\frac{0.65(1-0.65)}{100}}}\\p_{2}=0.65+(1.96\times 0.05)\\p_{1}=0.748\\p_{1}\approx0.75$