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3 years ago

7

When Andrei surveyed 36 random seventh-grade students in the lunchroom, he found that 7 out of 9 would like to try or have alrea

dy tried snorkeling. Using proportional reasoning, about how many of the 810 seventh-grade students in his school would like to snorkel?

2 answers:

Marysya12 [62]3 years ago

7 0

Answer:

630 people

Step-by-step explanation:

AnnyKZ [126]3 years ago

4 0

630 students would like to snorkel

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A soda bottling plant fills cans labeled to contain 12 ounces of soda. The filling machine varies and does not fill each can wit

AnnZ [28]

Answer:

$12.1-1.64\frac{0.35}{\sqrt{50}}=12.019$

$12.1+1.64\frac{0.35}{\sqrt{50}}=12.181$

So on this case the 90% confidence interval would be given by (12.019;12.181)

And we can conclude at 90% of confidence that the true mean is between (12.019, 12.181) and since the lower limit is higher than 12 we can conclude that the specification is not satisfied

Step-by-step explanation:

Previous concepts

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".

$\bar X=12.1$ represent the sample mean

$\mu$ population mean (variable of interest)

$\sigma=0.35$ represent the population standard deviation

n=50 represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.05,0,1)".And we see that $z_{\alpha/2}=1.64$

Now we have everything in order to replace into formula (1):

$12.1-1.64\frac{0.35}{\sqrt{50}}=12.019$

$12.1+1.64\frac{0.35}{\sqrt{50}}=12.181$

So on this case the 90% confidence interval would be given by (12.019;12.181)

And we can conclude at 90% of confidence that the true mean is between (12.019, 12.181) and since the lower limit is higher than 12 we can conclude that the specification is not satisfied

8 0

4 years ago

A large bank vault has several automatic burglar alarms. The probability is 0.55 that a single alarm will detect a burglar. (a)

Alborosie

Answer:

6 alarms to be 99% certain that a burglar trying to enter is detected by at least one alarm

Step-by-step explanation:

For each alarm, there are only two possible outcomes. Either a theft is detected, or is not. The probability of an alarm detecting a theft is independent of others alarms. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

0.55 that a single alarm will detect a burglar.

This means that $p = 0.55$

(a) How many such alarms should be used to be 99% certain that a burglar trying to enter is detected by at least one alarm?

The probability that no alarms detect a burglar is P(X = 0). Either no alarms detect, or at least one does. We need to have a 99% probability that at least one does. So we need P(X = 0) = 1 - 0.99 = 0.01.

We do by trial and error, find n until P(X = 0) <= 0.01.

n = 1

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{1,0}.(0.55)^{0}.(0.45)^{1} = 0.45$

n = 2

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{2,0}.(0.55)^{0}.(0.45)^{2} = 0.2025$

n = 3

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{3,0}.(0.55)^{0}.(0.45)^{3} = 0.091125$

n = 4

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{4,0}.(0.55)^{0}.(0.45)^{4} = 0.0410$

n = 5

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{5,0}.(0.55)^{0}.(0.45)^{5} = 0.0185$

n = 6

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{6,0}.(0.55)^{0}.(0.45)^{6} = 0.0083$

So we need 6 alarms to be 99% certain that a burglar trying to enter is detected by at least one alarm

8 0

3 years ago