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 <em>X</em> = number of senior professionals who thought that global warming is having a significant impact on the environment.
The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 100 and <em>p</em> = 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 <em>p</em> 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](https://tex.z-dn.net/?f=np%3D%20100%5Ctimes%200.65%3D65%3E10%5C%5Cn%281-p%29%3D100%5Ctimes%20%281-0.65%29%3D35%3E10)
Thus, a Normal approximation to binomial can be applied.
So,
.
(1)
Compute the value of
as follows:
![P(0.64](https://tex.z-dn.net/?f=P%280.64%3C%5Chat%20p%3C0.69%29%3DP%28%5Cfrac%7B0.64-0.65%7D%7B%5Csqrt%7B0.002275%7D%7D%3C%5Cfrac%7B%5Chat%20p-p%7D%7B%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D%7D%3C%5Cfrac%7B0.69-0.65%7D%7B%5Csqrt%7B0.002275%7D%7D%29)
![=P(-0.20](https://tex.z-dn.net/?f=%3DP%28-0.20%3CZ%3C0.80%29%5C%5C%3DP%28Z%3C0.80%29-P%28Z%3C-0.20%29%5C%5C%3D0.78814-0.42074%5C%5C%3D0.3674)
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
and
be the two population percentages that will contain the sample percentage with probability 90%.
That is,
![P(p_{1}](https://tex.z-dn.net/?f=P%28p_%7B1%7D%3C%5Chat%20p%3Cp_%7B2%7D%29%3D0.90)
Then,
![P(p_{1}](https://tex.z-dn.net/?f=P%28p_%7B1%7D%3C%5Chat%20p%3Cp_%7B2%7D%29%3D0.90)
![P(\frac{p_{1}-p}{\sqrt{\frac{p(1-p)}{n}}}](https://tex.z-dn.net/?f=P%28%5Cfrac%7Bp_%7B1%7D-p%7D%7B%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D%7D%3C%5Cfrac%7B%5Chat%20p-p%7D%7B%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D%7D%3C%5Cfrac%7Bp_%7B2%7D-p%7D%7B%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D%7D%29%3D0.90)
![P(-z](https://tex.z-dn.net/?f=P%28-z%3CZ%3Cz%29%3D0.90%5C%5CP%28Z%3Cz%29-%5B1-P%28Z%3Cz%29%5D%3D0.90%5C%5C2P%28Z%3Cz%29-1%3D0.90%5C%5C2P%28Z%3Cz%29%3D1.90%5C%5CP%28Z%3Cz%29%3D0.95)
The value of <em>z</em> for P (Z < z) = 0.95 is
<em>z</em> = 1.65.
Compute the value of
and
as follows:
![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](https://tex.z-dn.net/?f=z%3D%5Cfrac%7Bp_%7B2%7D-p%7D%7B%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D%7D%5C%5C1.65%3D%5Cfrac%7Bp_%7B2%7D-0.65%7D%7B%5Csqrt%7B%5Cfrac%7B0.65%281-0.65%29%7D%7B100%7D%7D%7D%5C%5Cp_%7B2%7D%3D0.65%2B%281.65%5Ctimes%200.05%29%5C%5Cp_%7B1%7D%3D0.7325%5C%5Cp_%7B1%7D%5Capprox0.73)
Thus, the two population percentages that will contain the sample percentage with probability 90% are 0.57 and 0.73.
(3)
Let
and
be the two population percentages that will contain the sample percentage with probability 95%.
That is,
![P(p_{1}](https://tex.z-dn.net/?f=P%28p_%7B1%7D%3C%5Chat%20p%3Cp_%7B2%7D%29%3D0.95)
Then,
![P(p_{1}](https://tex.z-dn.net/?f=P%28p_%7B1%7D%3C%5Chat%20p%3Cp_%7B2%7D%29%3D0.95)
![P(\frac{p_{1}-p}{\sqrt{\frac{p(1-p)}{n}}}](https://tex.z-dn.net/?f=P%28%5Cfrac%7Bp_%7B1%7D-p%7D%7B%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D%7D%3C%5Cfrac%7B%5Chat%20p-p%7D%7B%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D%7D%3C%5Cfrac%7Bp_%7B2%7D-p%7D%7B%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D%7D%29%3D0.95)
![P(-z](https://tex.z-dn.net/?f=P%28-z%3CZ%3Cz%29%3D0.95%5C%5CP%28Z%3Cz%29-%5B1-P%28Z%3Cz%29%5D%3D0.95%5C%5C2P%28Z%3Cz%29-1%3D0.95%5C%5C2P%28Z%3Cz%29%3D1.95%5C%5CP%28Z%3Cz%29%3D0.975)
The value of <em>z</em> for P (Z < z) = 0.975 is
<em>z</em> = 1.96.
Compute the value of
and
as follows:
![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](https://tex.z-dn.net/?f=z%3D%5Cfrac%7Bp_%7B2%7D-p%7D%7B%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D%7D%5C%5C1.96%3D%5Cfrac%7Bp_%7B2%7D-0.65%7D%7B%5Csqrt%7B%5Cfrac%7B0.65%281-0.65%29%7D%7B100%7D%7D%7D%5C%5Cp_%7B2%7D%3D0.65%2B%281.96%5Ctimes%200.05%29%5C%5Cp_%7B1%7D%3D0.748%5C%5Cp_%7B1%7D%5Capprox0.75)
Thus, the two population percentages that will contain the sample percentage with probability 95% are 0.55 and 0.75.