Sample statistics from uniform

5 The length, \(X \mathrm {~cm}\), of a piece of wooden planking is a random variable with probability density function given by $$f ( x ) = \begin{cases} \frac { 1 } { b } & 0 \leqslant x \leqslant b
0 & \text { otherwise } \end{cases}$$ where \(b\) is a positive constant.

1. Find the mean and variance of \(X\) in terms of \(b\). The lengths of a random sample of 100 pieces were measured and it was found that \(\Sigma x = 950\).
2. Show that the value of \(b\) estimated from this information is 19 . Using this value of \(b\),
3. find the probability that the length of a randomly chosen piece is greater than 11 cm ,
4. find the probability that the mean length of a random sample of 336 pieces is less than 9 cm .

2. The random variable \(X\) follows a continuous uniform distribution over the interval \([ \alpha - 3,2 \alpha + 3 ]\) where \(\alpha\) is a constant.
The mean of a random sample of size \(n\) is denoted by \(\bar { X }\)

1. Show that \(\bar { X }\) is a biased estimator of \(\alpha\), and state the bias. Given that \(Y = k \bar { X }\) is an unbiased estimator for \(\alpha\)
2. find the value of \(k\). A random sample of 10 values of \(X\) is taken and the results are as follows $$\begin{array} { l l l l l l l l l l } 3 & 5 & 8 & 12 & 4 & 13 & 10 & 8 & 5 & 12 \end{array}$$
3. Hence estimate the maximum value of \(X\)

8. The random variable \(X\) has a continuous uniform distribution over the interval \([ \alpha + 3,2 \alpha + 9 ]\) where \(\alpha\) is a constant. The mean of a random sample of size \(n\), taken from this distribution, is denoted by \(\bar { X }\)

1. Show that \(\bar { X }\) is a biased estimator of \(\alpha\)
2. Hence find the bias, in terms of \(\alpha\), when \(\bar { X }\) is used as an estimator of \(\alpha\) Given that \(Y = \frac { 2 \bar { X } } { 3 } + k\) is an unbiased estimator of \(\alpha\)
3. find the value of the constant \(k\) A random sample of 8 values of \(X\) is taken and the results are as follows
   4.8
   5.8
   6.5
   7.1
   8.2
   9.5
   9.9
   10.6
4. Use the sample to estimate the maximum value that \(X\) can take.
   \includegraphics[max width=\textwidth, alt={}]{3b6aaa8a-aeac-4a44-820b-e82f317d0c85-28_2633_1828_119_121}

4 The random variable \(X\) has a continuous uniform distribution on [ 0,10 ].

1. Find \(\mathrm { P } ( 3 < X < 6 )\).
2. Find each of the following.
   • \(\mathrm { E } ( X )\)
   • \(\operatorname { Var } ( X )\)
   Marisa is investigating the sample mean, \(Y\), of 8 independent values of \(X\). She designs a simulation shown in the spreadsheet in Fig. 4.1. Each of the 25 rows below the heading row consists of 8 values of \(X\) together with the value of \(Y\). All of the values in the spreadsheet have been rounded to 2 decimal places. \begin{table}[h]

   1	A	B	C	D	E	F	G	H	I	J
   1	\(X _ { 1 }\)	\(X _ { 2 }\)	\(X _ { 3 }\)	\(X _ { 4 }\)	\(X _ { 5 }\)	\(X _ { 6 }\)	\(X _ { 7 }\)	\(X _ { 8 }\)	\(Y\)
   2	6.31	2.45	3.27	3.06	4.16	1.53	0.43	7.99	3.65
   3	1.70	1.52	7.10	8.93	6.44	2.70	9.96	7.83	5.77
   4	9.15	0.52	4.95	6.99	6.52	3.15	0.81	5.35	4.68
   5	0.65	2.71	7.92	9.65	0.50	4.87	6.46	2.67	4.43
   6	3.09	6.11	3.96	0.09	0.18	4.67	0.67	6.20	3.12
   7	7.06	5.84	1.97	3.60	9.36	1.97	4.48	3.47	4.72
   8	1.46	1.57	5.45	0.37	3.76	7.56	8.48	9.12	4.72
   9	9.42	1.85	4.91	1.61	1.94	8.00	1.77	5.34	4.36
   10	2.98	5.32	2.91	4.12	9.16	1.76	9.97	6.88	5.39
   11	2.83	3.44	3.28	7.85	1.00	0.93	8.77	4.03	4.01
   12	4.51	0.59	5.84	9.87	8.65	3.94	7.18	0.23	5.10
   13	4.49	0.69	3.65	8.78	4.96	8.96	3.77	1.43	4.59
   14	6.57	8.08	4.85	6.75	7.92	0.27	9.69	4.04	6.02
   15	8.35	1.09	8.63	8.04	7.23	2.12	2.57	9.59	5.95
   16	5.24	9.53	6.08	8.21	3.61	7.07	6.65	7.63	6.75
   17	7.89	5.50	3.09	0.71	6.47	5.49	6.47	4.95	5.07
   18	8.36	7.27	2.35	9.04	0.58	2.26	3.01	7.90	5.10
   19	3.76	1.01	9.61	9.65	7.89	9.98	6.28	4.34	6.56
   20	9.94	6.84	3.38	5.53	0.26	8.53	5.72	5.12	5.66
   21	7.25	9.10	0.34	2.88	4.66	2.65	6.37	7.63	5.11
   22	7.18	7.14	5.38	0.04	4.09	6.47	4.96	4.23	4.94
   23	8.69	5.04	4.90	2.94	2.00	4.23	4.13	0.97	4.11
   24	3.46	6.33	0.48	9.35	0.23	1.18	7.97	6.37	4.42
   25	2.37	7.26	7.16	1.24	5.26	2.80	3.55	3.84	4.19
   26	2.16	8.30	7.17	3.32	2.96	1.30	9.11	0.31	4.33
   27

   \captionsetup{labelformat=empty} \caption{Fig. 4.1}
   \end{table}
3. Use the spreadsheet to estimate \(\mathrm { P } ( 3 < Y < 6 )\).
4. Explain why it is not surprising that this estimated probability is substantially greater than the value which you calculated in part (i). Marisa wonders whether, even though the sample size is only 8, use of the Central Limit Theorem will provide a good approximation to \(\mathrm { P } ( 3 < Y < 6 )\).
5. Calculate an estimate of \(\mathrm { P } ( 3 < Y < 6 )\) using the Central Limit Theorem. A Normal probability plot of the 25 simulated values of \(Y\) is shown in Fig. 4.2. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{0c58d4d7-10e9-473a-888a-b407ec90bf08-5_800_1291_306_386} \captionsetup{labelformat=empty} \caption{Fig. 4.2}
   \end{figure}
6. Explain what the Normal probability plot suggests about the use of the Central Limit Theorem to approximate \(\mathrm { P } ( 3 < Y < 6 )\). Marisa now decides to use a spreadsheet with 1000 rows below the heading row, rather than the 25 which she used in the initial simulation shown in Fig. 4.1. She uses a counter to count the number of values of \(Y\) between 3 and 6. This value is 808.
7. Explain whether the value 808 supports the suggestion that the Central Limit Theorem provides a good approximation to \(\mathrm { P } ( 3 < Y < 6 )\). Marisa decides to repeat each of her two simulations many times in order to investigate how variable the probability estimates are in each case.
8. Explain whether you would expect there to be more, the same or less variability in the probability estimates based on 1000 rows than in the probability estimates based on 25 rows.