Nom nom nom nom , . . . .

You find your dependent variable has a high degree of kurtosis. What is the first way to try to fix it?

Morgarella [4.7K]

The first way to try to fix this is to apply logarithm to the observations on the dependent variable. This is going to make the dependent variable with high degree of kurtosis normal.

Note that sometimes, the resulting values of the variable will be negative. Do not worry about this, as it is not a problem. It does not affect the regression coefficients, it only affects the regression intercept, which after transformation, will be of no interest.

7 0

3 years ago

Estimate the quotient for 49 / 311

Makovka662 [10]

To get a close estimate, we can round 49 up to 50 and 311 down to 300, obtaining an estimate of 50/300 = 1/6, or 0.1666... as a repeating decimal. That decimal approximation is a little less than one hundredth away from the actual decimal approximation of ≈ 0.1576

4 0

3 years ago

Read 2 more answers

What does the d stand for?

Zielflug [23.3K]

If you're talking about PEMDAS it's Division. (Parenthesis., Exponents, Multiplication, Division, Subtraction just in case you forget)

4 0

3 years ago

Which number has a prime factorization of 2x2x5x5x7x7

Dominik [7]

Answer:

4900

Step-by-step explanation:

multiply all those numbers together

2x2x5x5x7x7 = 4900

6 0

3 years ago

For a binomial distribution with p = 0.20 and n = 100, what is the probability of obtaining a score less than or equal to x = 12

notsponge [240]

The binomial distribution is given by,
P(X=x) = $(^{n}C_{x})p^{x} q^{n-x}$
q = probability of failure = 1-0.2 = 0.8
n = 100
They have asked to find the probability of obtaining a score less than or equal to 12.
∴ P(X≤12) = $(^{100}C_{x})(0.2)^{x} (0.8)^{100-x}$
where, x = 0,1,2,3,4,5,6,7,8,9,10,11,12
∴ P(X≤12) = $(^{100}C_{0})(0.2)^{0} (0.8)^{100-0} + (^{100}C_{1})(0.2)^{1} (0.8)^{100-1}$ + $(^{100}C_{2})(0.2)^{2} (0.8)^{100-2} + (^{100}C_{3})(0.2)^{3} (0.8)^{100-3}$ + $(^{100}C_{4})(0.2)^{4} (0.8)^{100-4} + (^{100}C_{5})(0.2)^{5} (0.8)^{100-5}$ + $(^{100}C_{6})(0.2)^{6} (0.8)^{100-6} + (^{100}C_{7})(0.2)^{7} (0.8)^{100-7}$ + $(^{100}C_{8})(0.2)^{8} (0.8)^{100-8} + (^{100}C_{9})(0.2)^{9} (0.8)^{100-9}$ + $(^{100}C_{10})(0.2)^{10} (0.8)^{100-10} + (^{100}C_{11})(0.2)^{11} (0.8)^{100-11}$ + $(^{100}C_{12})(0.2)^{12} (0.8)^{100-12}$

Evaluating each term and adding them you will get,
P(X≤12) = 0.02532833572
This is the required probability.

7 0

3 years ago