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makvit [3.9K]
2 years ago
6

The blood platelet counts of a group of women have a​ bell-shaped distribution with a mean of 247.9 and a standard deviation of

64.7. ​(All units are 1000 ​cells/μ​L.) Using the empirical​ rule, find each approximate percentage below. a. What is the approximate percentage of women with platelet counts within 2 standard deviations of the​ mean, or between 118.5 and 377.3​? b. What is the approximate percentage of women with platelet counts between 53.8 and 442.0​?

Mathematics
1 answer:
labwork [276]2 years ago
5 0

Answer:

A) Approximate percentage of women with platelet counts within 2 standard deviations of the​ mean, or between 118.5 and 377.3 = 95%

B) approximate percentage of women with platelet counts between 53.8 and 442.0 = 99.7%

Step-by-step explanation:

We are given;

mean;μ = 247.9

standard deviation;σ = 64.7

A) We want to find the approximate percentage of women with platelet counts within 2 standard deviations of the​ mean, or between 118.5 and 377.3.

Now, from the image attached, we can see that from the empirical curve, the probability of 1 standard deviation from the mean is (34% + 34%) = 68 %.

While probability of 2 standard deviations from the mean is (13.5% + 34% + 34% + 13.5%) = 95%

Thus, approximate percentage of women with platelet counts within 2 standard deviations of the​ mean, or between 118.5 and 377.3 = 95%

B) Now, we want to find the approximate percentage of women with platelet counts between 53.8 and 442.0.

53.8 and 442.0 represents 3 standard deviations from the mean.

Let's confirm that.

Since mean;μ = 247.9

standard deviation;σ = 64.7 ;

μ = 247.9

σ = 64.7

μ + 3σ = 247.9 + 3(64.7) = 442

Also;

μ - 3σ = 247.9 - 3(64.7) = 53.8

Again from the empirical curve attached, we cans that at 3 standard deviations from the mean, we have a percentage probability of;

(2.35% + 13.5% + 34% + 34% + 13.5% + 2.35%) = 99.7%

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Answer:

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b) \hat p=\frac{973}{1714}=0.568 represent the estimated proportion of people who showed partial or complete resistance to the antibiotic

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Step-by-step explanation:

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".

Part a

Description in words of the parameter p

p represent the real population proportion of people who showed partial or complete resistance to the antibiotic

\hat p represent the estimated proportion of people who showed partial or complete resistance to the antibiotic

n=1714 is the sample size required

z_{\alpha/2} represent the critical value for the margin of error

The population proportion have the following distribution  

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})

Part b

Numerical estimate for p

In order to estimate a proportion we use this formula:

\hat p =\frac{X}{n} where X represent the number of people with a characteristic and n the total sample size selected.

\hat p=\frac{973}{1714}=0.568 represent the estimated proportion of people who showed partial or complete resistance to the antibiotic

Part c

The confidence interval for a proportion is given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

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