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 households with incomes exceeding $38,000.

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

x -

median = 1.25

range = 12-14.5

interquartile = q12-q14.5

y -

median = 1

range = 15-17

interquartile = q15-q17

the answer is  

a.

28 ft2

step-by-step explanation: