The human cannonblall and other normal tortures
These Are Bob’s answers. I checked them and added some things. Since Bob failed to copyright his responses, I feel no guilt about stealing his work.
Thanks Bob!
The human cannon ball has flights that are normally distributed with an average of 150 feet and a standard deviation of 10 feet.
a. If his net is 30 feet long how often does he go to the hospital?
b. If he wants to go to the hospital less than 5% of the time how wide does his net have to be?
1. In Pennsylvania, wind speeds are distributed with a mean of 7.8 mph and a standard deviation of 19 mph. If telephone poles are designed to withstand the wind 99 percent of the time, how much wind must telephone poles be able to handle assuming wind velocity is normally distributed? Using the information given above, explain (prove) why wind is definitely not normally distributed.
.99
.01
.5 .49
0 7.8 x
Finding 0.4900 gives us a Z= 2.33. Therefore, the pole should withstand winds up to:
2.33 * 19 = 44.27 mph above mean, or 52 mph.
These data are terribly skewed positive, as 0 mph windspeed is as low as we go, but there is almost no limit to how high. Furthermore, the probability of negative wind if normal is : this is from excel using the normdist function. (fx) , statistical normdist
so 34 % of the time there should be negative wind, but we know this is impossible!
It can’t be normal.
4. If we assume that all taxpayers are honest, then the average amount that a person over or under pays her taxes should be zero. Half will overpay by mistake. The other half will underpay. Assume the standard deviation of the mistake is $500 and that the amount of the error is normally distributed. Each year the IRS randomly selects a number of taxpayers for an audit. Last year, Will Notpay was found to have underpaid his taxes by $1,257. He claimed, at the time, that it was an honest mistake. Based on the information provided, should the IRS believe him? Explain why or why not.
1257 is a Z = 2.51. This means that 99% of the people making errors will make an error smaller than Will Notpay. If we presume that the IRS truly did select him at random, Will should make prepare to become someone's Prison helper (sorry Bob, I had to alter your wording. This is a family show.) Using Excel the answer is:
There is a 99 percent chance that an honest mistake would be less than $1257 (less than 1% chance that the honest mistake would be over $1257. It seems that, Will is either dishonest or very unlucky.
2.
.99
.01
.5 .49
0 $1257
5. The recording space left unused at the end of a compact disk is normally distributed with an average of 3 minutes and a standard deviation of 1 minute. Whenever supergroup KISS records an album, they always have a 3.5 minute song ready just in case there is sufficient time at the end of the CD. How often will they get to use the extra song? Suppose they wanted to design a song that would fit at the end of the CD 95 percent of the time. How long should the song be?
SUPERGROUP?? Whom are you kidding?
The extra 0.5 minutes is z = 0.5; this means that 69% of the CDs will run shorter than this; they can add the extra song 31% of the time.
not enough time
enough time
.
3 3.5
As for the other question, finding 0.45 on the chart gives us a Z= 1.65. The song needs to be shorter than 3.0 - 1.65 = 1.35 minutes, which makes it a jingle.
not enough time
enough time .95
.05 .
.45 .5
x 3
6. The Electricity bills for PECO energy residential customers averages $25.29 for the month of April. The standard deviation for that month is $18.00. Assuming a normal distribution, what percent of customers will have a bill over $50.00? Using the information above, explain (prove) why April electricity bills can't be normally distributed.
If the bill is $50, this is Z= (50.00-25.29) / 18 = 1.37. Going to the chart, we find that Z= 1.37 yields a probability that the bill will be less than $50 91.47% of the time; the bill will exceed $50 8.5% of the time.
However, just like the wind example the bill cannot be less than zero, but it can certainly be much higher than 50. Positively skewed, and therefore not Normal.
Another approach is to recognize that a value of $50 (Z= 1.37) is exactly as likely as a bill for Z= - 1.37, or $0.58. This is ludicrous- PECO charges more than that to take your phone call to complain about your bill
How about a negative bill?:
so 8% of the time PECO owes you money?
7. Coffee consumption per person per day is normally distributed with an average of 16 oz and a standard deviation of 4 oz. What percentage of people consume more than 20 ounces per day? Design a cup that will hold enough coffee for 80 percent of the coffee drinkers.
Drink more than 20
.
16 20
20 oz is Z=1; 84.13% are less than or equal to 20, so 16% or so drink 20 or more.
For the second part, the chart gives us a Z= 0.84 for the 80th percentile (looking for .3 in the middle of the chart. A cup 19.36 oz would suffice. But so would 2 oz espresso cups, if the average person drank enough of them.
8.
80%
20%
.
50 30
16 x
9. In Pennsylvania, where the speed limit is 55 mph, the average speed of all cars on the road is 57.6 mph. Vehicle speed is normally distributed with a standard deviation of 8.3 mph. If a police officer wants to set his radar detector to sound the alarm when it records a driver is one of the fastest 5 percent, what speed should activate the alarm? Will Neverspeed was clocked at 30 mph in a 55 mph zone. He claimed he was moving with traffic. Based on a statistical analysis using the information provided, does this seem reasonable? Explain why or why not.
Using the chart to find 0.45, we get Z= 1.65. The man should set his radar at 1.65 * 8.3 = 13.7 mph above the average, or at 71.3 mph.
95%
5%
.
50 45
57.6 x
If Will was traveling at 30 mph, he was at Z= -3.33. This places Will going slower than 99.999 % of all traffic (my chart does not go to Z= 3.33). Will may have been going 30 mph, but the rest of traffic was not. I say we shoot him. (Golly Bob! maybe a ticket would be sufficient)