I included a screenshot with the question, thank you!!

According to a study conducted by the Gallup Organization, the the proportion of Americans who are afraid to fly is 0.10. A rand

notsponge [240]

Answer: 0.1457

Step-by-step explanation:

Let p be the population proportion.

Given: The proportion of Americans who are afraid to fly is 0.10.

i.e. p= 0.10

Sample size : n= 1100

Sample proportion of Americans who are afraid to fly = $\hat{p}=\dfrac{121}{1100}=0.11$

We assume that the population is normally distributed

Now, the probability that the sample proportion is more than 0.11:

$P(\hat{p}>0.11)=P(\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}>\dfrac{0.11-0.10}{\sqrt{\dfrac{0.10(0.90)}{1100}}})\\\\=P(z>\dfrac{0.01}{0.0090453})\ \ \ [\because z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}} ]\\\\=P(z>1.1055)\\\\=1-P(z\leq1.055)\\\\=1-0.8543=0.1457\ \ \ [\text{using z-table}]$

Hence, the probability that the sample proportion is more than 0.11 = 0.1457

3 0

3 years ago

Suppose that a standardized biology exam has a mean score of 80% correct, with a standard deviation of 3. The school administrat

NemiM [27]

Answer:

The 99% confidence interval is between 62.36%(lower bound) and 89.64%(upper bound).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.