(a) The density histogram above right represents a particular sample of n = 40 practice shots. Answer each of the following. Show all work.

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1 . Target Practice. An archer is practicing hitting the bull s-eye of the target shown below left. For any point on the target, define the continuous random variable D = (signed) radial distance to the center, as indicated by the dashed line. D (a) The density histogram above right represents a particular sample of n = 40 practice shots. Answer each of the following. Calculate the proportion (i.e., relative frequency) of shots that hit the solid inner circle. Recall that in a density histogram, the relative frequency corresponds to the area of the corresponding rectangle. Hence, Area = Density Width = (0.3)(2) = 0.6 (or 60%). SUM = Calculate the proportion (i.e., relative frequency) of shots that hit the middle band. Here, we have (0.5)() for 2 D, plus (0.5)() for D 2, = 0.3 (or 30%). Calculate the proportion (i.e., relative frequency) of shots that hit the outer band. Likewise, (0.05)() for 3 D 2, plus (0.05)() for 2 D 3, = 0. (or 0%). Determine the mean distance to the center, and its standard deviation (using the midpoints). Clearly, the sample mean is d =0 via symmetry. The frequencies f i are simply the given densities 5%, 5%, 60%, 5%, and 5% of n = 40, respectively, i.e., 2, 6, 24, 6, and 2. Thus, the sample variance is s = ( 2.5) (2) (.5) (6) (0) (24) (.5) (6) (2.5) (2) = 52/39 = 4/3 =.3333, so s =.3333, i.e., s =.54. For simplicity, we adopt the convention that positive values of D correspond to points on the right half of the target, and negative values of D correspond to points on the left half of the target, so that 3 D +3.

2 (b) Suppose that with continued practice, the true distribution of D is revealed to be approximately normal, with mean µ = 0, and standard deviation σ = ; that is, D ~ N(0, ), shown below left (to scale with the density histogram above). Answer each of the following. Calculate the probability of hitting the solid inner circle. D is just the standard normal distribution Z! Hence, P( D + ) = PZ ( + ) PZ ( ) = = Calculate the probability of hitting the middle band. By symmetry, P( 2 D ) + P( D 2) = 2 P( D 2) = [ PZ PZ ] [ ] 2 ( 2) ( ) = = 2(0.359) = Calculate the probability of hitting the outer band. Again, by symmetry, P( 3 D 2) + P(2 D 3) = 2 P(2 D 3) = [ PZ PZ ] [ ] 2 ( 3) ( 2) = = 2(0.0240) = [Note: This can alternatively be calculated as = Either is fine.] (c) Suppose the archer wants to improve his precision by reducing the standard deviation by 20%, to σ = ; see below right. What percentage of shots will have to hit the solid inner circle? P( D + ) = PD ( + ) PD ( ), but now D ~ N(0, ), so both + and will first need to be transformed to their corresponding z-scores. The formula X µ Z = yields σ 0 0 Z = =.25 and Z = =.25, respectively. Therefore, we now have PZ (.25) PZ (.25) = = 0.880, i.e., 8.8%.

3 2. The ages of a certain study population of children are normally distributed with mean µ = 600 days, and standard deviation σ = 200 days. (a) Calculate the probability that a child selected at random is one year old (i.e., 365 days) or less. 200 Let the population random variable A = Age ~ N(600, 200). We are A µ asked to find PA ( 365). Transform to the z-score via Z =, σ we obtain Z = =.5, which is exactly midway between 200 the z-table entries. and.8, so we may average the corresponding tabulated cumulative areas of 0.9 and 0.2, i.e., 0.2. Therefore, PA ( 365) = PZ (.5) = 0.2. (b) A sample of n = 264 children is to be randomly selected. Calculate the expected number and standard deviation of children who are one year old or less; assume independence between ages. (Hint: What is the distribution of the discrete random variable B = # children one year old or less in such a random sample?) The variable B is binomially distributed, i.e., B ~ Bin(264, 0.2) from part (a). Therefore, the mean (= expected value ) and standard deviation are, respectively, µ = nπ and σ = nπ( π). That is, µ = nπ = (264)(0.2) = 3.68 children, and σ = (264)(0.2)(8) = 5.28 children. (c) Calculate the exact probability that at least one child is one year old or less. 264 PB ( ) ( 0) (.2) (.88) = PB= = = (.88) 264 = (d) Calculate the probability that exactly 33 children are one year old or less P(B = 33) = (.2) (.88) = (e) Approximate (without continuity correction) the probability that 33 or fewer children are one year old or less Using the normal approximation to the binomial, we have via (b), Z = = 0.25, 5.28 so that PB ( 33) = PZ ( 0.25) = It is left as an exercise (though not required) to check that the conditions nπ 5 and n ( π ) 5 are satisfied.

4 3. As part of a nutritional questionnaire in a national health survey, individuals are asked the question How many days out of the past week did you eat at least one meal from a fast food chain restaurant (McDonald s, KFC, etc.)? The resulting population distribution of the response variable X turns out to be uniform, as shown in the probability histogram below. (a) Determine the mean number of days µ, and standard deviation σ for this population. 2 2 By symmetry, the mean is µ = 4 days. The variance is σ = ( x µ ) f( x) ( ) ( ) ( ) ( ) ( ) ( ) ( ) = ( 4) + (2 4) + (3 4) + (4 4) + (5 4) + (6 4) + ( 4) ( ) = + + = 4 days 2. Taking the square root yields the standard deviation σ = 2 days. (b) Suppose a sample of n = 36 individuals is to be randomly selected from the population. Calculate the approximate probability that the mean X is between 3.5 and 4.5 days. (9 pts) σ 2 From the Central Limit Theorem, it follows that X N µ, = N 4,, n 36 i.e., X N 4,. Therefore, to compute P 3 ( 3.5 X 4.5), it is first necessary to convert the endpoints to z-scores: The right endpoint becomes Z = =.5, and via 3 symmetry, the left becomes.5. Hence, we now have P(.5 Z +.5) = P( Z +.5) P( Z.5) = = (c) How large would a sample have to be, in order for there to exist a 9% probability that the mean number of days X is between 3.5 and 4.5 days? (0 pts) A central symmetric area of 0.9 in N(0, ) must have.05 in each tail, which corresponds to symmetric z-scores of ±2.. Hence, X 4 Z = ± 2. =. Using +2. with X = n (or 2. with X = 3.5) and solving for n yields n = (8.68) 2 = , so n 6.

5 4. Toxicity Testing. According to the EPA (Environmental Protection Agency), drinking water can contain no more than 0 ppb (parts per billion) of arsenic, in order to be considered safe for human consumption. Suppose that the concentration X of arsenic in a typical water source is known to be normally distributed, with an unknown mean µ and standard deviation σ. Since it is of no real concern if the mean µ is significantly below 0 ppb, only above, a one-sided hypothesis test is appropriate. In particular, we wish to test the left-tailed null hypothesis H : µ 0 0 ppb (i.e., safe) versus the right-tailed alternative H A: µ > 0 ppb (i.e., unsafe). (a) A random sample of n = 4 independent measurements yields a mean of x = 3.5 ppb and standard deviation s = 2.2 ppb. Calculate the one-sided p-value of this sample. Show all work! p-value = P( X 3.5) = PT 4 = PT ( ) = / 4 Based on your answer, fill out the following table. Significance Level Reject Null Hypothesis? (i.e., statistically significant?) Yes / No + reason Is there sufficient evidence to conclude that the water is unsafe? If Yes, is it weak, moderate, or strong? α =.05 Yes, since p =.025 <.05 Yes, moderate α =.0 No, since p =.025 >.0 No (b) Suppose a very similar random sample of n = 9 independent measurements yields a mean of x = 3.3 ppb and standard deviation s = 2.2 ppb. Calculate the one-sided p-value of this sample. Show all work! p-value = P( X 3.3) = PT 9 = PT ( 8 4.5) = / 9 Based on your answer, fill out the following table. Significance Level Reject Null Hypothesis? (i.e., statistically significant?) Yes / No + reason Is there sufficient evidence to conclude that the water is unsafe? If Yes, is it weak, moderate, or strong? α =.0 Yes, since p =.00 <<.0 Yes, strong (c) Based on your answers at the α =.0 level in parts (a) and (b), speculate why, IN GENERAL, two random samples with such a difference as these have, might lead to opposite conclusions. Clearly but briefly explain. (3 pts) Larger sample size yields greater power to detect statistical significance, if present. magic word This is known as the Maximum Contaminant Level (MCL).