A recent report indicated a typical family of four spends $490 per month on food
ID: 3171747 • Letter: A
Question
A recent report indicated a typical family of four spends $490 per month on food. Assume the distribution of food expenditures for a family of four follows the normal distribution, with a standard deviation of $90 per month. (Round your intermediate answers to 2 decimal places. Round your answers to 2 decimal places.)
A recent report indicated a typical family of four spends $490 per month on food. Assume the distribution of food expenditures for a family of four follows the normal distribution, with a standard deviation of $90 per month. (Round your intermediate answers to 2 decimal places. Round your answers to 2 decimal places.)
Explanation / Answer
Mean ( u ) =490
Standard Deviation ( sd )=90
Normal Distribution = Z= X- u / sd ~ N(0,1)
a.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 30) = (30-490)/90
= -460/90 = -5.1111
= P ( Z <-5.1111) From Standard Normal Table
= 0
P(X < 490) = (490-490)/90
= 0/90 = 0
= P ( Z <0) From Standard Normal Table
= 0.5
P(30 < X < 490) = 0.5-0 = 0.5
b.
P(X < 430) = (430-490)/90
= -60/90= -0.6667
= P ( Z <-0.6667) From Standard Normal Table
= 0.2525 = 25.25% are spend less than 430
c.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 430) = (430-490)/90
= -60/90 = -0.6667
= P ( Z <-0.6667) From Standard Normal Table
= 0.25249
P(X < 600) = (600-490)/90
= 110/90 = 1.2222
= P ( Z <1.2222) From Standard Normal Table
= 0.88919
P(430 < X < 600) = 0.88919-0.25249 = 0.6367 ~ 63.67% are b/w 430 to 600
d.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 500) = (500-490)/90
= 10/90 = 0.1111
= P ( Z <0.1111) From Standard Normal Table
= 0.54424
P(X < 600) = (600-490)/90
= 110/90 = 1.2222
= P ( Z <1.2222) From Standard Normal Table
= 0.88919
P(500 < X < 600) = 0.88919-0.54424 = 0.345 ~ 34.5% are b/w 500 to 600
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