(a) If <em>f(x)</em> is to be a proper density function, then its integral over the given support must evaulate to 1:
For the integral, substitute <em>u</em> = <em>x</em> ² and d<em>u</em> = 2<em>x</em> d<em>x</em>. Then as <em>x</em> → 0, <em>u</em> → 0; as <em>x</em> → ∞, <em>u</em> → ∞:
which reduces to
<em>c</em> / 2 (0 + 1) = 1 → <em>c</em> = 2
(b) Find the probability P(1 < <em>X </em>< 3) by integrating the density function over [1, 3] (I'll omit the steps because it's the same process as in (a)):
We can solve these equations by using substitution method where you substitute one equation to the other and solve for the value of the other variable. We do as follows:
y = 2x^2
y = –3x −1
2x^2 = –3x −1
2x^2 +3x +1 = 0
x1= -0.5
x2 = -1
y1 = 0.5
y2 = 2
Answer:
a) P(X≥143)=0
b) This contradicts the study as getting a sample with this proportion is almost impossible (if the proportion of 68% is true).
Step-by-step explanation:
If we use the normal approximation to the binomial distribution we have the following parameters (mean and standard deviation):
Then, we can calculate the probability of X being equal or more than 143 using the z-score:
This contradicts the study as getting a sample with this proportion is almost impossible (if the proportion of 68% is true).
