Encounters between cruise ships and whales are perhaps inevitable. but researchers for the National Park Service are hoping to implement strategies to reduce these encounters. In 2009. observers for the National Park Service recorded over 1,500 unique encounters between whales and cruise ships in the Glacier Bay National Park and Preserve. In 20% of these encounters. the whale came within 3,231 feet ofthe bow of the ship. representing a "near miss".1 'Gende. S. M. Harris. FL, Nielsen. J.. 8i Hendrix. A. N. {n.d.). Using;r observers to record encounters between cruise ships and hurnlobock wholes. National Park Service. https:mvw.nps.govra rticlesfaps-9-2~4.htrn Suppose another team of researchers in 2009 believed that the rate of encounters in which the whale comes within 3,281 feet of the bow in the lower bay sub-region of Glacier Bay was higher than 20%. This team observed a sample of 85 encounters between cruise ships and whales in the lower bay: the whale came within 3,281 feet of the bow in 25 of these encounters. Pa rt A: Assumingp = 0.20. is it valid to use the normal approximation to the sampling distribution of, the sample proportion of encounters where the whale came within 3,281 feet of the bow? justify your answer. -,since85{ll2l]) = 17 3 ' land 35(1 0.20) = I38 10. Pa rt B: Assuming p = 0.20. calculate the mean and standard deviation ofj for samples of size 85. Mean = Standard deviation [rounded to four decimal places] = - Pa rt C: Assumingp = 0.20, use the normal distribution to approximate the probability that the whale came within 3,281 feet in 25 or more of a sample of 85 encounters. Round to the nearest thousandth. Pa rt D: Based on your answer in Part C, do these data provide evidence that the rate by which a whale came within 3,281 feet in the lower bay was higher than the overall rate of 20%?justify your answer. . Ifthe rate in the lower bay was 28%. we would only see data like ours in about I |% ofall samples of 85 encounters. Since these data are so to occur under the assumption that the rate was 20%. this suggests the rate was actually than 20%. Question Help: E Message instructor E Calculator Submit