In a poll of 2000 likely voters, 960 say that the US spends too little on fighting hunger at home. Find a 96% confidence interval for the true proportion of voters who feel this way. a. Find a point estimate for p, the population proportion who thought US spends too little on fighting hunger at home. b.Find a 96% confidence interval for p.

a. 0.48

b. The 96% confidence interval for p is (0.457, 0.503).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of , we have the following confidence interval of proportions.

In which

z is the zscore that has a pvalue of .

In a poll of 2000 likely voters, 960 say that the US spends too little on fighting hunger at home.

This means that .

This is the answer for question a.

96% confidence level

So , z is the value of Z that has a pvalue of , so .

The lower limit of this interval is:

The upper limit of this interval is:

The 96% confidence interval for p is (0.457, 0.503).

percentage - 25%

probability - 0.25

we know,

total amount of students in class = 20

total boys = 10

total girls = 10

no of girls who wear glasses = 10/2 = 5

probability of choosing a girl with glasses is =

number of girls wearing glasses/total number of students in the class = 5/20

thus, percentage of selecting a girl wearing glasses will be

= 5/20 × 100

= 5 × 5

25%

therefore,

probability will be 0.25

happy to

pls mark as brainliest

9 x 8 divided by (5-3)=36