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Art [367]
3 years ago
5

The population of a city on April 15, 1915, was 47,175. During the period between March 1 and July 1, 1915, 1,325 new cases of s

mallpox occurred in the city. Of these cases, 642 persons were ill with smallpox according to surveillance reports on June 1, 1915. The monthly incident rate (per 1,000) of active cases of influence for the 4-month period was:
Mathematics
1 answer:
Norma-Jean [14]3 years ago
7 0

Answer:

the incidence rate is 28.47

Step-by-step explanation:

The computation of the monthly incident rate of active cases is shown below:

Incidence rate is

= 1325 ÷  (47175 - 642) × 1000

= (1325 ÷ 46533 ) ×  1000

= 28.47

hence, the incidence rate is 28.47

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Suppose that birth weights are normally distributed with a mean of 3466 grams and a standard deviation of 546 grams. Babies weig
Anon25 [30]

Answer:

3.84% probability that it has a low birth weight

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 3466, \sigma = 546

If we randomly select a baby, what is the probability that it has a low birth weight?

This is the pvalue of Z when X = 2500. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{2500 - 3466}{546}

Z = -1.77

Z = -1.77 has a pvalue of 0.0384

3.84% probability that it has a low birth weight

3 0
3 years ago
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