The price of one share of stock in the Pilsdorff Beer Company (see [[guide:Ee45340c30#sec 8.2 |Exercise.]]) is given by [math]Y_n[/math] on the [math]n[/math]th day of the year. Finn observes that the differences [math]X_n = Y_{n + 1} - Y_n[/math] appear to be independent random variables with a common distribution having mean [math]\mu = 0[/math] and variance [math]\sigma^2 = 1/4[/math]. If [math]Y_1 = 100[/math], estimate the probability that [math]Y_{365}[/math] is
- [math]{} \geq 100[/math].
- [math]{} \geq 110[/math].
- [math]{} \geq 120[/math].
Test your conclusions in Exercise by computer simulation. First choose 364 numbers [math]X_i[/math] with density [math]f(x) = \mbox {normal}(x,0,1/4)[/math]. Now form the sum [math]Y_{365} = 100 + X_1 + X_2 +\cdots+ X_{364}[/math], and repeat this experiment 200 times. Make up a bar graph on [math][50,150][/math] of the results, superimposing the graph of the approximating normal density. What does this graph say about your answers in Exercise?
Physicists say that particles in a long tube are constantly moving back and forth along the tube, each with a velocity [math]V_k[/math] (in cm/sec) at any given moment that is normally distributed, with mean [math]\mu = 0[/math] and variance [math]\sigma^2 = 1[/math]. Suppose there are [math]10^{20}[/math] particles in the tube.
- Find the mean and variance of the average velocity of the particles.
- What is the probability that the average velocity is [math]{} \geq 10^{-9}[/math] cm/sec?
An astronomer makes [math]n[/math] measurements of the distance between Jupiter and a particular one of its moons. Experience with the instruments used leads her to believe that for the proper units the measurements will be normally distributed with mean [math]d[/math], the true distance, and variance 16. She performs a series of [math]n[/math] measurements. Let
be the average of these measurements.
- Show that
[[math]] P\left(A_n - \frac 8{\sqrt n} \leq d \leq A_n + \frac 8{\sqrt n}\right) \approx .95. [[/math]]
- When nine measurements were taken, the average of the distances turned out to be 23.2 units. Putting the observed values in (a) gives the 95 percent confidence interval for the unknown distance [math]d[/math]. Compute this interval.
- Why not say in (b) more simply that the probability is .95 that the value of [math]d[/math] lies in the computed confidence interval?
- What changes would you make in the above procedure if you wanted to compute a 99~percent confidence interval?