Suppose that insurance companies did a survey. They randomly surveyed 410 drivers and found that 300 claimed they always buckle up. We are interested in the population proportion of drivers who claim they always buckle up.
NOTE: If you are using a Student's t-distribution, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)
(i) Enter an exact number as an integer, fraction, or decimal.
x =
(ii) Enter an exact number as an integer, fraction, or decimal.
n =
(iii) Round your answer to four decimal places.
p' =
Which distribution should you use for this problem? (Round your answer to four decimal places.)
P' _ ( , )

x = 300

n = 410

p' = 0.7317

Step-by-step explanation:

From the given information;

the objective is to answer the following:

(i) Enter an exact number as an integer, fraction, or decimal.

Mean x = 300

(ii) Enter an exact number as an integer, fraction, or decimal.

Sample size n = 410

(iii) Round your answer to four decimal places.

Sample proportion p' of the drivers who always claimed they buckle up is :

p' = x/n

p' = 300/410

p' = 0.7317

Which distribution should you use for this problem? (Round your answer to four decimal places.)

P' _ ( , )

The normal distribution is required to be used because we are interested in proportions and the sample size is large.

Let consider X to be the random variable that follows a normal distribution.

X represent the number of people that always claim they buckle up

∴

d

step-by-step explanation:

d

sorry, i don't know i just need points, i'm new here. chao