A standard number cube has 6 sides....
probability of rolling a 5 is 1/6......probability of not rolling a 5 is 5/6
Step-by-step explanation:
just substitute the value of g(X) and f(X)
the answer of this question is 6×10 6
Answer:
0.150,0.595
Step-by-step explanation:
Given that at a self-service gas station, 40% of customers pump regular gas, 35% pump midgrade, and 25% pump premium gas. Of those who pump regular, 30% pay at least $30. Of those who pump midgrade, 50% pay at least $30. And of those who pump premium, 60% pay at least $30.
Regular gas Midgrade Premium gas Total
Percent 40 35 25 100
atleast 30 30% 50% 60%
a) The probability that the next customer pumps premium gas and pays at least $30
=
b) the probability that the next customer pays at least $30
= P(regular and pays atleast 30%)+P(premium and pays atleast 30%)+P(midgrade and pays atleast 30%)
=
Answer:
Step-by-step explanation:
Hello!
For me, the first step to any statistics exercise is to determine what is the variable of interest and it's distribution.
In this example the variable is:
X: height of a college student. (cm)
There is no information about the variable distribution. To estimate the population mean you need a variable with at least a normal distribution since the mean is a parameter of it.
The option you have is to apply the Central Limit Theorem.
The central limit theorem states that if you have a population with probability function f(X;μ,δ²) from which a random sample of size n is selected. Then the distribution of the sample mean tends to the normal distribution with mean μ and variance δ²/n when the sample size tends to infinity.
As a rule, a sample of size greater than or equal to 30 is considered sufficient to apply the theorem and use the approximation.
The sample size in this exercise is n=50 so we can apply the theorem and approximate the distribution of the sample mean to normal:
X[bar]~~N(μ;σ2/n)
Thanks to this approximation you can use an approximation of the standard normal to calculate the confidence interval:
98% CI
1 - α: 0.98
⇒α: 0.02
α/2: 0.01
X[bar] ±
174.5 ±
[172.22; 176.78]
With a confidence level of 98%, you'd expect that the true average height of college students will be contained in the interval [172.22; 176.78].
I hope it helps!