stat400-hw07-Sp24-soln

Solution Let X = alcohol consumption in grams where X ~ Normal(57, 18 2 ). Then x 0.75 is the value such that P( X ≤ x 0.75 ) = 0.75. P ሺ𝑋 ൑ 𝑥 ଴ . ଻ହ ሻ ൌ 0.75 ⇒ P ሺ𝑍 ൑ 𝑧 ଴ . ଻ହ ሻ ൌ 0.75 ⇒ P ሺ𝑍 ൑ 0.68 ሻ ൌ 0.75 ⇒ 𝑧 ଴ . ଻ହ ൌ 0.68 𝑥 ଴ . ଻ହ ൌ μ ൅ 𝑧 ∙ σ ൌ 57 ൅ 0.68 ∙ 18 ൌ 69.24 grams Exercise 2 A chocolatier produces caramel-filled chocolates that have a labeled weight of 20.4 grams. Assume that the distribution of the weights of these caramel-filled chocolates is N (21.37, 0.16). (a) Let X denote the weight of a single chocolate selected at random from the production line. Find P ( X > 22.07). Solution Let X = weight of a single chocolate where X ~ Normal(21.37, 0.16). P ሺ X ൐ 22.07 ሻ ൌ P ൬ 𝑋 െ 21.37 √ 0.16 ൐ 22.07 െ 21.37 √ 0.16 ൰ ൌ P ሺ𝑍 ൐ 1.75 ሻ ൌ P ሺ𝑍 ൏ െ 1.75 ሻ ൌ 0.0401 (b) Suppose that 15 caramel-filled chocolates are selected independently and weighed. Let Y equal the number of these chocolates that weigh less than 20.857 grams. Find P ( Y ≤ 2). Solution Let X = weight of a single chocolate where X ~ Normal(21.37, 0.16). Let Y = number of chocolates weighing less than 20.857 out of 15 total, where Y ~ Binomial(15, p = P( X < 20.857)) P ሺ X ൏ 22.07 ሻ ൌ P ൬ 𝑋 െ 21.37 √ 0.16 ൏ 20.857 െ 21.37 √ 0.16 ൰ ൌ P ሺ𝑍 ൏ െ 1.28 ሻ ൌ 0.1003, 𝑡ℎ𝑜𝑢𝑔ℎ 𝑖𝑡 𝑤𝑜𝑢𝑙𝑑 𝑏𝑒 𝑜𝑘𝑎𝑦 𝑡𝑜 𝑟𝑜𝑢𝑛𝑑 𝑡𝑜 0.10. Thus, Y ~ Binomial(15, 0.10).