(a) Calculate the bee s mean final position on the hexagon, and clearly label this position on the figure below. Show all work.

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35, or independently, along the right-hand edge with probability 0.65, and repeats the process, until she has completed a total of n = 6 moves.

1 1. A worker bee inspects a hexagonal honeycomb cell, starting at corner A. When done, she proceeds to an adjacent corner (always facing inward as shown), either by randomly moving along the lefthand edge with probability 0.35, or independently, along the right-hand edge with probability 0.65, and repeats the process, until she has completed a total of n = 6 moves. (a) Calculate the bee s mean final position on the hexagon, and clearly label this position on the figure below. Show all work. Solution: Let L = # moves to the left; then R = 6 L = # moves to the right. We are told that the corresponding probabilities of Success and Failure are π = 0.35 and π = 0.65, respectively. Therefore, since each move is independent of any other, the bee s entire path forms a sequence of Bernoulli trials; hence both L and R are binomially distributed. That is, L ~ Bin(6, 0.35) and R ~ Bin(6, 0.65), so that mean # moves to the left = µ L = nπ L = (6)(0.35) = 2.1, and likewise, mean # moves to the right = µ R = nπ R = (6)(0.65) = 3.9, which places the mean position of the bee at = 1.8 moves to the right, as shown below. L R (b) How many ways are there for her to end up at corner C? Solution: The bee can only end up at C if, out of a total of n = 6 moves, either L = 1 (and hence R = 5), or L = 4 (and hence R = 2). Since, for each possibility, the order of moves does not affect the outcome, it follows that there are ways that L = 1, and ways that L = 4. So the total 1 4 number is + = 21 ways. 1 4 (c) Set up BUT DO NOT EVALUATE an expression for the probability that she ends up back where she started (A). Show all work. Solution: The bee can only end up back at A if, out of a total of n = 6 moves, either L = 0 (and hence R = 6), or L = 3 (and hence R = 3), or L = 6 (and hence R = 0). Via the binomial probability formula, we therefore have the following. C D X E P(L = 0) = B F P(L = 3) = A P(L = 6) = Thus, the answer is (0.65) + (0.35) (0.65) + (0.35). 3

2 2. The National Weather Service (NWS) estimates that the probability of being killed by lightning in the United States with 300 million individuals is 1 in 5 million per year. (a) In a single year, what is the expected number of lightning fatalities in the U.S.? Show all work. (2 pts) Solution: This can be modeled using either the Binomial distribution or, because it is a rare 1 event, the Poisson distribution. Either way, with n = 300 million and π =, we have 5 million the mean µ (or λ) = nπ = (300)(1/5) = 60 expected cases per year. (b) Set up AND NUMERICALLY EVALUATE an expression for the probability that there are exactly 65 fatalities per year. Show all work. (8 pts) Solution: Let X = # lightning fatalities per year. Binomial: P(X = 65) = Poisson: P(X = 65) = 300,000,000 ( ) ( ) e (60) 65! ,999,935 or, much simpler, The Poisson calculation is clearly preferable, and yields a probability. (c) The NWS suspects that, due to under-reporting, the actual mean number of lightning fatalities is closer to 70 per year. Using this information, set up BUT DO NOT EVALUATE an expression for the probability that there are exactly 65 fatalities per year. Show all work. 70 Solution: In this case, nπ = 70, so that π = 300 million, i.e, π = , 000, ,999,935 Binomial: P(X = 65) = π (1 π) 65 or, much simpler, e (70) Poisson: P(X = 65) = 65!

3 3. In a certain urban population, annual salary X (in $K) is normally distributed, with mean µ = 65, and standard deviation = 20. Answer each of the following; show all work! X ~ N (65, 20) = 20 µ = 65 (a) What salary marks the cutoff for the lowest 20% of the population? Solution: From the Z-table, we see that the z-score corresponding to a cumulative probability of 0.2 is z = Hence, Z = yields X = 65 + ( 0.84)(20) = $48.2K. (b) What salary marks the cutoff for the highest 10% of the population? Solution: An upper tail probability of 0.1 is equivalent to a lower tail probability of 0.9. From the Z-table, we see that the z-score corresponding to a cumulative probability of 0.9 is z = Hence, Z = yields X = 65 + (1.28)(20) = $90.6K. (c) Calculate the probability that a randomly selected individual earns over $100K. Solution: Via Z =, the z-score corresponding to X = 100 is Hence, PX ( > 100) = PZ ( > 1.75) = = Z = = (d) Calculate the probability that a randomly selected individual earns over $100K, given that he/she earns over $65K. PX ( > 100) Solution: PX ( > 100 X> 65) = = = PX ( > 65) 0.5

4 (e) Suppose a sample of n = 4 individuals is randomly selected from the population. Calculate the probability that their mean salary is over $100K. Solution: From the preliminary statement that preceded the Central Limit Theorem, we know that if the population distribution X ~ N ( µ, ), then the sampling distribution X ~ N µ,, for any sample size n. Thus, in this problem, the standard error is n 20 = = 10, and so X ~ N (65,10). Therefore, via Z =, the z-score n corresponding to X = 100 is Z = = Hence, PX ( > 100) = PZ ( > 3.5) = =

4. In a supermarket population of mixed nuts, 32% are 6-oz bags, 36% are 12-oz bags, and the remaining 32% are 18-oz bags. Answer each of the following; show all work!

5 4. In a supermarket population of mixed nuts, 32% are 6-oz bags, 36% are 12-oz bags, and the remaining 32% are 18-oz bags. Answer each of the following; show all work! (a) Sketch a probability histogram of this distribution. Label all relevant features (b) Compute the mean and standard deviation (in oz) of this distribution. Show all work. Solution: By symmetry, it is clear that the mean is µ = 12 oz. [Alternatively, µ = xf( x) = (6)(.032) + (12)(0.36) + (18)(0.32) = 12. ] The variance is 2 2 = ( x µ ) f( x) = (6 12) (0.32) + (12 12) (0.36) + (18 12) (0.32) = 23.04, so the standard deviation is = 4.8 oz. (c) Calculate the approximate probability that the mean weight of a random sample of n = 36 bags is between 11 oz and 13 oz. Justify all claims! Show all work. Solution: The population distribution has finite mean and variance, and in addition, is symmetric. The sample size is greater than 30, so the Central Limit Theorem certainly applies: Via X N µ,. The standard error is n 4.8 = = 0.8, and so X N(12, 0.8). n 36 Z =, the z-scores corresponding to X = 13 and X = 11 are ± 1 Z = = ± 1.25, 0.8 respectively. Thus, P(11 X 13) = P( 1.25 Z ) = = (d) How large would a random sample of bags have to be, in order to guarantee an approximate 95% probability that the mean weight is between 11 oz and 13 oz? Show all work. Solution: The symmetric z-scores ±1.96 contain 95% probability between them. Therefore, Z = yields ± 1 ± 1.96 =, i.e., n = ( ) 2 = 88.51, so n 89 bags. 4.8 / n

Probability Distributions.

Probability Distributions. Probability Distributions http://www.pelagicos.net/classes_biometry_fa18.htm Probability Measuring Discrete Outcomes Plotting probabilities for discrete outcomes: 0.6 0.5 0.4 0.3 0.2 0.1 NOTE: Area within