The reported unemployment is 5.5% of the population. what measurement scale is used to measure unemployment? nominal ordinal int
1 answer:
You might be interested in
Answer:
Step-by-step explanation:
Hello!
You have the variable
X: Area that can be painted with a can of spray paint (feet²)
The variable has a normal distribution with mean μ= 25 feet² and standard deviation δ= 3 feet²
since the variable has a normal distribution, you have to convert it to standard normal distribution to be able to use the tabulated accumulated probabilities.
a.
P(X>27)
First step is to standardize the value of X using Z= (X-μ)/ δ ~N(0;1)
P(Z>(27-25)/3)
P(Z>0.67)
Now that you have the corresponding Z value you can look for it in the table, but since tha table has probabilities of , you have to do the following conervertion:
P(Z>0.67)= 1 - P(Z≤0.67)= 1 - 0.74857= 0.25143
b.
There was a sample of 20 cans taken and you need to calculate the probability of painting on average an area of 540 feet².
The sample mean has the same distribution as the variable it is ariginated from, but it's variability is affected by the sample size, so it has a normal distribution with parameners:
X[bar]~N(μ;δ²/n)
So the Z you have to use to standardize the value of the sample mean is Z=(X[bar]-μ)/(δ/√n)~N(0;1)
To paint 540 feet² using 20 cans you have to paint around 540/20= 27 feet² per can.
c.
P(X≤27) = P(Z≤(27-25)/(3/√20))= P(Z≤2.98)= 0.999
d.
No. If the distribution is skewed and not normal, you cannot use the normal distribution to calculate the probabilities. You could use the central limit theorem to approximate the sampling distribution to normal if the sample size was 30 or grater but this is not the case.
I hope it helps!
Answer:
78% probability that a randomly selected online customer does not live within 50 miles of a physical store.
Step-by-step explanation:
A probability is the number of desired outcomes divided by the number of total outcomes.
In this problem, we have that:
Total outcomes:
100 customers
Desired outcomes:
A clothing vendor estimates that 78 out of every 100 of its online customers do not live within 50 miles of one of its physical stores. So the number of desired outcomes is 78 customers.
Using this estimate, what is the probability that a randomly selected online customer does not live within 50 miles of a physical store?
78% probability that a randomly selected online customer does not live within 50 miles of a physical store.