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3 years ago

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A pair of dice is rolled, and the number that appears uppermost on each die is observed. Refer to this experiment and find the p

robability of the given event. (Enter your answer as a fraction.) One die shows a 5, and the other is a number less than 4.

1 answer:

nikitadnepr [17]3 years ago

8 0

1 Is the answer because you have to subtract 5 - 4 equals 1. did that help?

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Listed below are systolic blood pressure measurements (mm Hg) taken from the right and left arms of the same woman. Consider the

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Answer:

a) $\bar d= \frac{\sum_{i=1}^n d_i}{n}=72.2$

$s_d =\sqrt{\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1}} =9.311$

b) The 90% confidence interval would be given by (63.330;81.070)

$63.330 < \mu_{left}- \mu_{right}$

Step-by-step explanation:

1) Previous concepts and notation

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".

Let put some notation

x=left arm , y = right arm

x: 175 169 182 146 144

y: 102 101 94 79 79

The first step is define the difference $d_i=x_i-y_i$ , that is given so we have:

d: 73, 68, 88, 67, 65

The second step is calculate the mean difference

$\bar d= \frac{\sum_{i=1}^n d_i}{n}=72.2$

The third step would be calculate the standard deviation for the differences, and we got:

$s_d =\sqrt{\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1}} =9.311$

2) Confidence interval

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

$\bar d \pm t_{\alpha/2}\frac{s_d}{\sqrt{n}}$ (1)

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 0.90 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: "=-T.INV(0.05,4)".And we see that $t_{\alpha/2}=2.13$

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

$72.2-2.13\frac{9.311}{\sqrt{5}}=63.330$

$72.2+2.13\frac{9.311}{\sqrt{5}}=81.070$

So on this case the 90% confidence interval would be given by (63.330;81.070)

$63.330 < \mu_{left}- \mu_{right}$

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