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Alexxandr [17]
3 years ago
5

The average diameter of sand dollars on a certain island is 5.00 centimeters with a standard deviation of 0.90 centimeters. If 1

6 sand dollars are chosen at random for a collection, find the probability that the average diameter of those sand dollars is more than 4.73 centimeters. Assume that the variable is normally distributed.
Mathematics
1 answer:
Montano1993 [528]3 years ago
4 0

Answer:

the probability that the average diameter of those sand dollars is more than 4.73 centimeters is 0.5910

Step-by-step explanation:

Given information:

mean, x = 5

standard deviation, σ = 0.9

number of sample, n = 16

now we calculate the probability that the average diameter of those sand dollars is more than 4.73 centimeters.

P(x>4.73)  = P (z > (4.73 - x) / (σ \sqrt{n})

                = P (z > (4.73 - 4) / (0.9 \sqrt{16})

                = P (z > 0.203)

                = 0.5910

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Valentin [98]

Answer:

\bar X= \sum_{i=1}^n \frac{x_i}{n} (2)  

s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}} (3)  

The mean calculated for this case is \bar X=12.8

The sample deviation calculated s=4.324 \approx 4.3

12.8-4.604\frac{4.324}{\sqrt{5}}=3.896    

12.8+ 4.604\frac{4.324}{\sqrt{5}}=21.704    

So on this case the 99% confidence interval would be given by (3.896;21.704)    

Step-by-step explanation:

Data: 10,9,20,13,12

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 represent the sample mean for the sample  

\mu population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

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

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

In order to calculate the mean and the sample deviation we can use the following formulas:  

\bar X= \sum_{i=1}^n \frac{x_i}{n} (2)  

s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}} (3)  

The mean calculated for this case is \bar X=12.8

The sample deviation calculated s=4.324

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=5-1=4

Since the Confidence is 0.99 or 99%, the value of \alpha=0.01 and \alpha/2 =0.005, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,4)".And we see that t_{\alpha/2}=4.604

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

12.8-4.604\frac{4.324}{\sqrt{5}}=3.896    

12.8+ 4.604\frac{4.324}{\sqrt{5}}=21.704    

So on this case the 99% confidence interval would be given by (3.896;21.704)    

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Alchen [17]
<h3>Answer:  -3/4  (choice C)</h3>

==================================================

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