Ricky has found that for every belt buckle he sells, he makes 7 + bx 4 dollars in profit, where b is the number of belt
1 answer:
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Answer:
4.72 hours/day
Step-by-step explanation:
Mean time spent watching TV (μ) = 2.8 hours a day
Standard deviation (σ) = 1.5 hours a day
The 90th percentile (upper 10%) of a normal distribution has an equivalent z-score of roughly z = 1.282. The minimum time spent watching TV, X, at the 90th percentile is:
On a typical day, you must watch at least 4.72 hours of TV to be in the upper 10%.
We assume the lunch prices we observe are drawn from a normal distribution with true mean and standard deviation 0.68 in dollars.
We average samples to get
.
The standard deviation of the average (an experiment where we collect 45 samples and average them) is the square root of n times smaller than than the standard deviation of the individual samples. We'll write
Our goal is to come up with a confidence interval (a,b) that we can be 90% sure contains .
Our interval takes the form of as
is our best guess at the middle of the interval. We have to find the z that gives us 90% of the area of the bell in the "middle".
Since we're given the standard deviation of the true distribution we don't need a t distribution or anything like that. n=45 is big enough (more than 30 or so) that we can substitute the normal distribution for the t distribution anyway.
Usually the questioner is nice enough to ask for a 95% confidence interval, which by the 68-95-99.7 rule is plus or minus two sigma. Here it's a bit less; we have to look it up.
With the right table or computer we find z that corresponds to a probability p=.90 the integral of the unit normal from -z to z. Unfortunately these tables come in various flavors and we have to convert the probability to suit. Sometimes that's a one sided probability from zero to z. That would be an area aka probability of 0.45 from 0 to z (the "body") or a probability of 0.05 from z to infinity (the "tail"). Often the table is the integral of the bell from -infinity to positive z, so we'd have to find p=0.95 in that table. We know that the answer would be z=2 if our original p had been 95% so we expect a number a bit less than 2, a smaller number of standard deviations to include a bit less of the probability.
We find z=1.65 in the typical table has p=.95 from -infinity to z. So our 90% confidence interval is
in other words a margin of error of
dollars
That's around plus or minus 17 cents.