Question

1.9% of mosquitos caught at random in any given field study carry malaria. If you conduct a study and capture 72 mosquitos, how many mosquitos infected with malaria would you expect to have in your mosquito-catching Mason jar out of all 72 mosquitos buzzing around? Round up to the nearest whole number if your calculation results in a number with decimal point values. For example, 0.2 rounds up to 1 and 74.25 rounds up to 75.

2. A research study examined the blood vitamin D levels of the entire US population of landscape gardeners. The population average level of vitamin D in US landscapers was found to be 42.85 ng/mL with a standard deviation of 4.034 ng/mL. Assuming the true distribution of blood vitamin D levels follows a Gaussian distribution, what is the upper value of the region for which approximately 68% of the data are located within the distribution? Recall the 68-95-99.9% observation discussed in the "Normal Distribution" section of the textbook that is helpful in answering the question.

Answer #1

We would be looking at the first question here:

Q1) We are given here that:

P( malaria ) = p = 0.09

n = 72 is the sample size here.

The number of mosquitoes we would expect to be infected with malaria that we would expect to have in our mosquito-catching Mason jar out of all 72 mosquitos buzzing around would be computed here as:

= np

= 72*0.09

= 6.48

Therefore 6.48 is the expected number of mosquitoes here.

**Therefore 7 is the required rounded up +1 value
here.**

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PLEASE DO NOT ROUND YOUR ANSWER. THANKS
A research study examined the blood vitamin D levels of the
entire US population of landscape gardeners. The population average
level of vitamin D in US landscapers was found to be 554 ng/mL with
a standard deviation of 4.191 ng/mL. Assuming the true distribution
of blood vitamin D levels follows a normal distribution, if you
randomly select a landscaper in the US, what is the probability
that his/her vitamin D level will be...

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