Question

Q1. Suppose at random 40% of school children develop nausea and vomiting following holiday parties and that you conduct a study to examine this phenomenon, with a sample size of n=50. What is the probability that 12 or more children become sick?

Answer:

Q2. 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 52.07 ng/mL with a standard deviation of 4.567 ng/mL. Assuming the true distribution of blood vitamin D levels follows a Gaussian distribution, what is the 66-th percentile? Recall that the statistic, "percentile", is defined to be the value below which a given percentage of observations are located within the distribution. In other words, for this question find the vitamin D value for which 66% of the observations within the distribution are located to the left of this value.

Answer:

Answer 1

1. Let: X be a random variable denoting the number of children develop nausea and vomiting following holiday parties among 50 children.

Now, 40% of school children develop nausea and vomiting following holiday parties.
So, the probability that one child develop nausea and vomiting is = 0.40.

Clearly,  

The pmf of X is given by:

x = 0 , 1 , 2 , .............. , 50.

Now, we've to calculate the probability that 12 or more children become sick.

The probability that 12 or more children become sick is given by = 0.99431.

2. Let: Y be the true distribution of blood vitamin D levels follows a Gaussian distribution, i.e., Normal Distribution.
i.e.,

Since, we are to calculate the 66^th percentile, i.e., the point below which 66% of the data lie is equivalent to find the value of y for which F(y) = 0.66. where F(.) is the c.d.f. of the random variable, Y.

Therefore, the  the vitamin D value for which 66% of the observations within the distribution are located to the left of this value is 53.9537

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