Either my brain is fried or i’m missing something here, but I cant seem to figure out how the answer to this question is reached: A food retailer has determined that the mean household income of her customers is $47 500 with a standard deviation of $12 500. She is trying to justify carrying a line of luxury food items that woud appeal to households with incomes greater than $60 000. Based on her information and assuming that household incomes are normally distributed, what percentage of households in her customer base has incomes of $60 000 or more? A) 5.00% B) 2.50% C) 15.87% D) 34.13% Now the answer is C), with the explanation given as: From the table of areas under the normal curve, 34.13% of observations lie outside 1 standard deviation of the mean. So 50% - 34.13% =15.8% with incomes of $60 000 or more. I cant seem to make sense of the explantion or i’m reading the table wrong. I calculated the Z-stat to be 1, and from there im stuck. Please help anyone???
Zstat = 1 Look up 1.00 on the ztable, you get .8413. That means 84.13 will be less than 60,000. (1-.8413) = .1587 so 15.87% will be above. At least this is my reasoning
Thanks, i’m only seeing it more clearly now, think there might be a typo in the explanation that was given. The .8143 also make more sense to me, couldnt figure why or how they were using 34.13% but its most probably a mistake.
notice that 60,000 = 47,500 + 12,500 = mean + 1*stdev. Roughly 67% is contained between mean - stdev and mean + stdev -> in the right tail: (1-67%)/2 = 33%/2 = 16.5% -> C
you have to find the Z score (Actual - Expected)\ SD = 47.5 - 60 / 12.5 = 1 look up the Z score for 1 and it equals .68, which you should know anyway. 1-.68 = .32% total but you need to find only 6 or greater so you divide by 2 = .32/2 = .16%=15.87
The incomes are normally distributed so you have to account for both ends of the curve. Therefore 60,000 = 47500 + (z x 12500) and 35,000 = 47500 - (z x 12500) z = +1 z = -1 We know that being +/- 1 standard deviation away from the mean give a confidence interval of about 68 percent. Because the z charts are the cumulative probability you get a value of 0.8413, but because you have to take to account your lower bound which has a cumulative prob of 0.1587. .6826 = .8413 - .1587 If you draw the bell curve anything to the right of 60,000 will give you 0.1587 chance of being greater than that number and anything to the left (or less than) 35,000 will also have a probability of 0.1587. The Z tables are cumulative so you have to subtract left tail in order to get a proper confidence interval. Hopefully that helps and I didn’t confuse you. D
why are you guys making this so complicated? It’s a basic Equation. Calc Z score: (Actual - expected) / standard deviation = Z Score. just plug in the numbers like I did above 47.5 - 60 / 12.5 = 1 U should know that a z score of 1 = .68, just like 1.645 = 90% and 1.96 = 95% so on and so on. If u don’t know then look it up on the Z Score chart 1-.68%= .32%, which represents both tailes you need only one tail, the 60k or greater so u divide by 2 and it equals 16%, closest answer C. Can’t get any easier than that.
Thanks guys, its much clearer now. Now I gotta go fry my brain some more 
so is my reasoning wrong above?
no you were fine.
You don’t even need to calculate anything You know 60K is 1 Stdev greater than the mean value of 47500. This is provided, there isn’t a need to calculate anything. Approx 68% of the possible values will lie within +/- 1 std dev. This means 32% lie outside. Since you only care about one of the tails, divide 32/2 and you get 16. This is all approximate so the closest answer is C.