The household income in a community is normally distributed with a mean of $42,000 and a STDev of $5,000. Find the proportion of
1 answer:
Answer:
The proportion of households with incomes exceeding $38.000 is 0.788
Step-by-step explanation:
Define X your random variable ( a variable whose outcome is probabilistic or depending in chance) as:
X - Household income in community which is normally distributed or N(42,000. 5000) (mean and STDV).
To answer the question we most use the cumulative distribution function of the normal distribution (CDF). The CDF tells you the probability that your random variable X is less than a value or P(X<a) where is any value. In the problem a is 38.000 but they ask you for P(X>38,000). Following probability rules for continuous variables this is the same as P(X>38,000)= 1 - P(X<38,000).
As there is no explicit formula for the CDF of a normal variable, we use the standard normal distribution which is a normal distribution with mean 1 and STDev 0 - value expressed usually as Z. To standarize our random nromal variable we subtract the mean and divide by the standard deviation of our random variable X in this case:
then you can replace your X variable for the standard Z variable
Then we can use the Z table or Excel to find our proportion as follows.
P(Z>38000)=1-P(Z<38000)
You can find this probabilty in Excel Norm.Dist(-0.8,42000,5000, True) or using the Z table. In the rows you look for the value of the probability "-0.8") and in the columns for the value of the hundredths of your probability in this case is 0. This gives you a value of 0,211855399 . So:
1-P(Z<-0.80)=1-0,211855399=0,788144601 that rounded is 0.788
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Answer:
a)TRUE
b)FALSE
c)TRUE
d)FALSE
e)TRUE
f)TRUE
g)TRUE
h)FALSE
i)FALSE
j)TRUE
Step-by-step explanation:
a) For every x there is y such that :
TRUE
This statement is true, because for every real number there is a square number of that number, and that square number is also a real number. For example, if we take 6.5, there is a square of that number and it equals 39.0625.
b) For every x there is y such that :
FALSE
For example, if x = -1, there is no such real number so that its square equals -1.
c) There is x for every y such that xy = 0
TRUE
If we put x = 0, then for every y it will be xy=0*y=0
d)There are x and y such that
FALSE
There are no such numbers. If we rewrite the equation we obtain an incorrect statement:
e)For every x, if there is y such that xy=1:
TRUE
The statement is true. If we have a number x, then multiplying x with 1/x (Since x is not equal to 0 we can do this for ever real number) gives 1 as a result.
f)There is x for every y such that if then xy=1.
TRUE
The statement is equivalent to the statement in e)
g)For every x there is y such that x+y = 1
TRUE
The statement says that for every real number x there is a real number y such that x+y = 1, i.e. y = 1-x
So, the statement says that for every real umber there is a real number that is equal to 1-that number
h) There are x and y such that
We have to solve this system of equations.
From the first equation it yields x=2-2y and inserting that into the second equation we have:
Which is obviously false statement, so there are no such x and y that satisfy the equations.
FALSE
i)For every x there is y such that
We have to solve this system of equations.
From the first equation it yields and inserting that into the second equation we obtain:
Inserting that back to the first equation we obtain
So, there is an unique solution to this equations:
x=1 and y=1
The statement is FALSE, because only for x=1 (and not for every x) exists y (y=1) such that
j)For every x and y there is a z such that
TRUE
The statament is true for all real numbers, we can always find such z. z is a number that is halway from x and from y.