Question

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' _ ( , )

Answer 1

Answer:

x = 300

n = 410

p' = 0.7317

$\mathbf{P' \sim Normal (\mu = 0.7317, \sigma = 0.02188)}$

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

∴

$P' \sim Normal (\mu = p' , \sigma = \sqrt{\dfrac{p(1-p)}{n}})$

$P' \sim Normal (\mu = 0.7317, \sigma = \sqrt{\dfrac{0.7317(1-0.7317)}{410}})$

$P' \sim Normal (\mu = 0.7317, \sigma = \sqrt{\dfrac{0.7317(0.2683)}{410}})$

$P' \sim Normal (\mu = 0.7317, \sigma = \sqrt{\dfrac{0.19631511}{410}})$

$P' \sim Normal (\mu = 0.7317, \sigma = \sqrt{4.78817341*10^{-4}})$

$\mathbf{P' \sim Normal (\mu = 0.7317, \sigma = 0.02188)}$