Answers when I get back this afternoon.
- A tire manufacturer produces a particular model tire whose tread wear life is normally distributed with a mean of 39,000 miles and a standard deviation of 5,300 miles. The manufacturer wishes to provide a guaranteed tread life for this model which would be exceeded by 98% of all tires. What tread life would meet this requirement?
- The mechanical process which fills 10-lb bags of dog food is subject to random fluctuations in the amount placed in each bag. The amount placed in each bag is approximately normally distributed with a mean of 170 ounces and a standard deviation of 4.3 ounces. Determine an interval centered on the mean such that the weight of the contents of 99% of the bags will fall within that interval.
- The scores on an exam are approximately normally distributed with a mean of 75 and a standard deviation of 10. If the professor wants 10% of the class to receive As, then what is the minimum score a student can get and receive an A on the exam?
Tire? I don't need anyone to manufacture tiredness for me, I already have that in abundance. Has the 'y' gone from the American version?
ReplyDeleteIf we're not doing metric then I'll sulk.
I'll keep my deviancy to myself as long as you give me an A.
Cheers
It's tire here, and we haven't yet thrown away our cultural heritage in favor of what the French do, and gone metric.
ReplyDeleteThank God.
Apparently no one else considers this math fun to be fun math. Oh well. :-)
ReplyDeleteLooks like 2 sig figs all around, which meshes nicely with my method of Windows calc.exe and trusty CRC book containing tables of erf(x) (actually F(x), which is more convenient).
Disclaimer: I haven't had statistics, and I haven't been a student in 20+ years. Plus, I'm doing these on my lunch break instead of eating my lunch, so who knows if I'm doing them right. Here goes anyway:
Let F(x) be the cdf, i.e. the area from -inf to x under the probability density f(x) of the standardized (centered at 0) normally distributed (which we've been told these are) random variable.
Since the standard normal distribution is symmetric and centered at 0, we can use the CRC tables and the relationship F(-x) = 1 - F(x). The x that we come up with will be the number of standard deviations.
1. "exceeded by 98% of all" indicates that we are looking for the point where the area under the curve to the right of x, from x to infinity, is 98% or 0.98, which means that the area from 0 to x is 0.02, i.e. F(x) = 0.02. The table only gives values where F(x) > 0.5, so we have to use the F(-x) = 1 - F(x) relationship. 1 - F(x) = 0.98, which falls between F(2.05) = 0.9798 and F(2.06) = 0.9803. Extrapolating linearly (close enough), we say that 0.98 ~= F(2.054). 0.02 = 1 - 0.98 ~= 1 - F(2.054) = F(-2.054). Therefore, we are looking at x ~= 2.054 s.d. to the left of the mean. Mean = 39,000, s.d. = 5300 => answer = 39,000 - 2.054(5300) = 28113.8
answer 1: 28,000 miles (to 2 sig figs)
2. "interval centered on the mean such that ... 99% ... will fall within that interval" If 0.99 fall within the interval (-x, x), then 0.99/2 = 0.495 fall within the interval (0, x), and 0.495 + 0.5 = 0.995 fall within the interval (-inf, x). This is F(x). The table gives F(2.57) = 0.9949 and F(2.58) = 0.9951. Extrapolating, we have 0.995 ~= F(2.575). So x ~= 2.575 s.d. and our interval is 170 oz +/- 2.575(4.3) oz, or 170 oz +/- 11.0725 oz.
answer 2: 160-180 oz. (to 2 sig figs)
3. "10% of the class to receive As" (and assume that A is the highest grade) means that we are looking for x where F(x) = 0.90. The table gives F(1.28) = 0.8997 and F(1.29) = 0.9015. Extrapolating, we have F(1.2817) ~= 0.90. Mean = 75, s.d. = 10, so we are looking for 75 + 10(1.2817) = 87.817.
answer 3: 88 (to 2 sig figs)