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Serga [27]
3 years ago
11

Hope you can read those plz help me

Mathematics
1 answer:
Annette [7]3 years ago
6 0

the answer to the first question is B

the answer to the second question is also B i think


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Answer pls<br> 3278+32+3213+89321+31398
goldenfox [79]

Answer:

oop

Step-by-step explanation:

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Evaluate the expression for n=-3
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5n is your answer hope this help

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A sample survey is designed to estimate the proportion of sports utility vehicles being driven in the state of California. A ran
mart [117]

Answer:

a) The 95% confidence interval would be given (0.070;0.121).

b) "increase the sample size n"

"decrease za/2 by decreasing the confidence"

Step-by-step explanation:

Notation and definitions

X=48 number of vehicles classified as sports utility.

n=500 random sample taken

\hat p=\frac{48}{500}=0.096 estimated proportion of vehicles classified as sports utility vehicles.

p true population proportion of vehicles classified as sports utility vehicles.

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})

(a) Use a 95% confidence interval to estimate the proportion of sports utility vehicles in California. (Round your answers to three decimal places.)

The confidence interval would be given by this formula :

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 95% confidence interval the value of \alpha=1-0.5=0.05 and \alpha/2=0.025, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=1.96

And replacing into the confidence interval formula we got:

0.096 - 1.96 \sqrt{\frac{0.096(1-0.096)}{500}}=0.070

0.096 + 1.96 \sqrt{\frac{0.096(1-0.096)}{500}}=0.121

And the 95% confidence interval would be given (0.070;0.121).

We are confident (95%) that about 7.0% to 12.1% of vehicles in California are classified as sports utility .  

(b) How can you estimate the proportion of sports utility vehicles in California with a higher degree of accuracy? (HINT: There are two answers. Select all that apply.)

For this case we just have two ways to increase the accuracy one is "increase the sample size n" since if we have a larger sample size the estimation would be more accurate. And the other possibility is "decrease za/2 by decreasing the confidence" because if we decrease the confidence level the interval would be narrower and accurate

7 0
3 years ago
How do you classify a triangle by its sides
erik [133]
Triangle classification by sides would be scalene, no congruent sides; isosceles, at least 2 congruent sides; equilateral, 3 congruent sides.
6 0
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The times of the runners in a marathon are normally distributed, with a mean of 3 hours and 50 minutes and a standard deviation
serious [3.7K]

The probability that a randomly selected runner has a time less than or equal to 3 hours and 20 minutes is 0.16 or 16%.

<h3>What is normally distributed data?</h3>

Normally distributed data is the distribution of probability which is symmetric about the mean.

The mean of the data is the average value of the given data.

The standard deviation of the data is the half of the difference of the highest value and mean of the data set.

The times of the runners in a marathon are normally distributed, with

  • Mean of 3 hours and 50 minutes
  • Standard deviation of 30 minutes.

Refere the probabiliity table attached below. The probability of Z being inside the 1 Standard daviation of mean is 0.84.

The probability of runner selected with time less than or equal to 3 hours and 20 minutes,

P=1-0.84\\P=0.16

Thus, the probability that a randomly selected runner has a time less than or equal to 3 hours and 20 minutes is 0.16 or 16%.

Learn more about the normally distributed data here;

brainly.com/question/6587992

4 0
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