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dimulka [17.4K]
3 years ago
14

Of the volunteers coming into a blood center, 1 in 3 have O+ blood, 1 in 15 have O-, 1 in 3 have A+, and 1 in 16 have A-. The na

me of one person who previously has donated blood is selected from the records of the center. What is the probability that the person selected has________.a. type O+ bloodb. type O bloodc. type A bloodd. neither type A nor type O blood
Mathematics
1 answer:
Zanzabum3 years ago
8 0

Answer:

Step-by-step explanation:

Given that of  the volunteers coming into a blood center, 1 in 3 have O+ blood, 1 in 15 have O-, 1 in 3 have A+, and 1 in 16 have A-.

i.e. we can say out of the volunteers

O+ = \frac{1}{3} = 0.3333

O-=\frac{1}{15}=0.06667

A+=\frac{1}{3}=0.3333

A-=\frac{1}{16}=0.0625

Total = 0.795833

We can say remaining belong to B or AB.

%he probability that the person selected has________.

a. type O+ blood  = \frac{0.3333}{0.7958} =0.4188

b. type O blood= \frac{0.3333+0.06667}{0.7958} =0.5026

c. type A blood = \frac{0.3333+0.0625}{0.7958} =0.4974

d. neither type A nor type O blood = 1-0.7958\\= 0.2042

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5 0
3 years ago
The head librarian at the Library of Congress has asked her assistant for an interval estimate of the mean number of books check
Verdich [7]

Answer:

n=(\frac{1.960(150)}{90})^2 =10.67 \approx 11

So the answer for this case would be n=11 rounded up to the nearest integer

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 represent the sample mean for the sample  

\mu population mean (variable of interest)

\sigma =150 represent the population 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)

The confidence interval for this case is given by: (740, 920)

We can find the estimate for the mean and we got:

\bar X = \frac{740+920}{2} = 830

and the margin of error is given by :

ME = \frac{920-740}{2}= 90

The margin of error is given by this formula:

ME=z_{\alpha/2}\frac{s}{\sqrt{n}}    (a)

And on this case we have that ME =90 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

n=(\frac{z_{\alpha/2} s}{ME})^2   (b)

The critical value for 95% of confidence interval now can be founded using the normal distribution. And in excel we can use this formla to find it:"=-NORM.INV(0.025;0;1)", and we got z_{\alpha/2}=1.960, replacing into formula (b) we got:

n=(\frac{1.960(150)}{90})^2 =10.67 \approx 11

So the answer for this case would be n=11 rounded up to the nearest integer

5 0
3 years ago
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