Answer:
(a) 3.75
(b) 2.00083
(c) 0.4898
Step-by-step explanation:
It is provided that X has a continuous uniform distribution over the interval [1.3, 6.2].
(a)
Compute the mean of X as follows:

(b)
Compute the variance of X as follows:

(c)
Compute the value of P(X < 3.7) as follows:
![P(X < 3.7)=\int\limits^{3.7}_{1.3}{\frac{1}{6.2-1.3}}\, dx\\\\=\frac{1}{4.9}\times [x]^{3.7}_{1.3}\\\\=\frac{3.7-1.3}{4.9}\\\\\approx 0.4898](https://tex.z-dn.net/?f=P%28X%20%3C%203.7%29%3D%5Cint%5Climits%5E%7B3.7%7D_%7B1.3%7D%7B%5Cfrac%7B1%7D%7B6.2-1.3%7D%7D%5C%2C%20dx%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B4.9%7D%5Ctimes%20%5Bx%5D%5E%7B3.7%7D_%7B1.3%7D%5C%5C%5C%5C%3D%5Cfrac%7B3.7-1.3%7D%7B4.9%7D%5C%5C%5C%5C%5Capprox%200.4898)
Thus, the value of P(X < 3.7) is 0.4898.
I honestly don't understand
Answer:
0.0903
Step-by-step explanation:
Given that :
The mean = 1450
The standard deviation = 220
sample mean = 1560



P(X> 1560) = P(Z > 0.5)
P(X> 1560) = 1 - P(Z < 0.5)
From the z tables;
P(X> 1560) = 1 - 0.6915
P(X> 1560) = 0.3085
Let consider the given number of weeks = 52
Mean
= np = 52 × 0.3085 = 16.042
The standard deviation =
The standard deviation = 
The standard deviation = 3.3306
Let Y be a random variable that proceeds in a binomial distribution, which denotes the number of weeks in a year that exceeds $1560.
Then;
Pr ( Y > 20) = P( z > 20)


From z tables
P(Y > 20)
0.0903