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NeTakaya
3 years ago
13

Dion practiced the long jump . He jumped 3.05 meters,2.74 meters and 3.3 meters.What was the average length of Dion's jumps?

Mathematics
2 answers:
Brums [2.3K]3 years ago
7 0
The average of the lengths of Dion's jump is 3.03 meters
Mademuasel [1]3 years ago
5 0
All you do is subtract 3.05 and 2.74 then do the same thing with the other ones and see what number is between all of them. What grade are you in
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A sample of eight workers in a clothing manufacturing company gave the following figures for the amount of time(in minutes) need
lilavasa [31]

Answer:

10.108 < \mu < 13.892    

Step-by-step explanation:

1) 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  

We have the following distribution for the random variable:

X \sim N(\mu , \sigma=0.45)

And by the central theorem we know that the distribution for the sample mean is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

2) Confidence interval

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

The sample deviation calculated s=2.268

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

df=n-1=8-1=7

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-T.INV(0.025,7)".And we see that t_{\alpha/2}=2.36

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

12-2.36\frac{2.268}{\sqrt{8}}=10.108    

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So on this case the 95% confidence interval would be given by (10.108;13.892)

10.108 < \mu < 13.892    

7 0
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victus00 [196]

Answer:

The answer to this question is B

Step-by-step explanation:

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