25 Math 175 worksheet #5 Name. Show all work on these Canvas using 1 pdf file. D

Question

25 Math 175 worksheet #5 Name. Show all work on these Canvas using 1 pdf file. DO NOT SUBMIT MULTIPLE FILES! #1.) In a recent year, Delaware had the highest per capita annual income with $51.80 with ? $4850. (a) If (b) If 2 (c) If a resident is randomly selected, what is the probability that their income is more than $53,000? Define a variable, draw and shade a bell, solve, and summarize pages to receive full credit. When finished, scan and upload these pages onto 16 residents are randomly selected, what is the probability that the mean income less than $49,0002 5 residents are randomly selected, what is the probability that the mean income is more than $53,000

Explanation / Answer

1) mu=51800 and sigma = 4850

a) n=16 Xbar = 49000

P(Xbar<49000) = ? is the asked question

We need to use the z value formula

z = (xbar-mu)/(sigma/sqrt(n)) = (49000-51800)/ (4850/sqrt(16)) = -2.309278

Since we are looking for P(Xbar<49000) we need to find the area under the z curve for z value < -2.309278

So from the z table we get area under the curve = 0.01046407

and hence we conclude that 0.01046407 is the probability of mean income less than 49000

b)Same as shown above n=25 , Xbar=53000 and P(Xbar>53000)

We need to use the z value formula

z = (Xbar-mu)/(sigma/sqrt(n)) = (53000-51800)/ (4850/sqrt(25)) = 1.237113

Since we are looking for P(Xbar>53000) we need to find the area under the z curve for z value > 1.237113

So from the z table we get area under the curve = 0.1080225

and hence we conclude that 0.1080225is the probability of mean income more than 53000

c)Same as shown above n=1 (as we are now selecting 1 resident randomly), Xbar=53000 and P(Xbar>53000)

We need to use the z value formula

z = (Xbar-mu)/(sigma/sqrt(n)) = (53000-51800)/ (4850/sqrt(1)) = 0.2474227

Since we are looking for P(Xbar>53000) we need to find the area under the z curve for z value > 0.2474227

So from the z table we get area under the curve = 0.4022906

and hence we conclude that 0.4022906 the probability of mean income more than 53000

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