Worksheet #2 Mean amount for Genuardis Market is $27.21

Worksheet #2 Mean amount for Genuardis Market is $27.21

Worksheet
#2 Mean amount for Genuardis Market is $27.21, Standard deviation of $7.93
with a roughly normal distribution.
a. The middle 95% spent between what two amounts? Since this distribution is approximately normal, using the 68-95-97.5 rule
approximately 95% of the middle shoppers are between 2 standard deviations
of the mean. So 27.71 + 2*7.63 and 27.21- 2*7.63. Therefore the middle 95%
of the customers spent between $12.45 and $42.97.
Either answer is acceptable as long as you justify your method If you choose to use the calculator or table:
invnorm(.975, 27.21, 7.93)
Yields $42.75. And. Invnorm(.025,27.21,7.93)
Yields $11.67
b. Person spends $10, find percentile.
You can use the calculator
NormCdf(- 100000 too hard to find infinity symbol, 10,27.21,7.93)
Yields 0.014994
or use z = (10-27.21)/7.93 z=-2.17 P(z -2.17)=.015
A person who spends $10 at the Genuardis Markets on a typical Saturday
would be at approximately the 1.5%.
c. Person spending $40 is in the top 5.5% of spending on a typical Saturday.
Using calculator NormCdf(40, 1000000, 27.21, 7.93) Yields .053387
P(z 1.61) = .053 Using z score formula and table.
d. What range of amounts do the middle 60% represent?
invnorm (.80,27.21,7.93)
yields 33.8841
invnorm(.20, 27.21, 7.93)
yields 20.5359
The middle 60% of saturday shoppers at the G.....Market typically spend between
$20.54 and $33.88.
e. spends at top 2%
invnorm(0.98,27.21,7.93)
yields 43.4962
A customer who is the top
2% of
spending on a typical Saturday night at G...... Market spends at least $43.50.
f. invnorm(.995, 27.21, 7.93) yields 47.6363 A customer who spends at the 99.5 percentile on the typical Saturday night
spends about $47.63.
g. 99.99th percentile
invnorm(.9999, 27.21,7.93) yields
56.7018 A customer who spends at the 99.99 percentile spends approximately
$56.70 on a typical Saturday night at the G..... Market.
f. b. a. Calculator: invnorm(.025, 200,52)
this yields 98.0819
invnorm (.975, 200,52) this yields 301.918
The middle 95% of the drivers at the busy intersection spend between 98.1
seconds and 301.9 seconds. b. Exactly 2 minutes which is 120 seconds.
normCdf(-10000000, 120,200,52)
yields 0.061968
If a driver waits exactly 2 minutes at the busy intersection , this would be at the
6th percentile.
c. If a person waits 6 minutes he is in the top 0.1 percent of waiting.
invnorm(360, 10000000,200,52) yields .001046 d. normCdf(120, 240, 200, 52) yields .717154
e. Driver waits for top 2%
invnorm(0.98,200,52) yields 306.795
If a driver waits in the top 2%, then he waits approximately 306.8 seconds.
f. If a customer spends in the 0.5 percentile of waiting times, then he spends
approximately 66.06 seconds.
Invnorm(0.005,200,52) yields 66.0569