Question

women have head circumferences that are normally distributed with a mean given by u-21.78 in, and a standard deviation given by ơ:06 in. Complete parts a through c below a. If a hat company produces women's hats so that they fit head circumferences between 21.3 in. and 22.3 in, what is the probability that a randomly selected woman will be able to fit into one of these hats? The probability is Round to four decimal places as needed)

Answer 1

a = 21.3 NotElement b = 22.3: To find: p(a < x < b) given: mu = 21.78 NotElement sigma = 0.6. p(a < x < b) = p(21.3 - 21.78/0.6 < z < 22.3 - 21.78/0.6) = p(-0.8 < z < 0.866) = 0.8069 - 0.2118 = 0.5951 or 59.51% (Note: Area under Normal distribution curve upto, 21.3 is 0.248 NotElement upto 22.3 is 0.8069] mu = 0.8539g: sigma = 0.0512: a = 0.8535 Therefore p(a < x) = p(0.8535 - 0.8539/0.0512 < z) = p(-0.0078 < z). = 0.4968 or 49.68% Therefore required probability = 100 - 49.68 = 50.32% mu = 183.3: sigma = 40.7: a = 169. Therefore p(a < x) = p(169 - 183.3/40.7 < z) = p(-0.3514 < z) = 0.3627 or 36.27% Therefore required probability = 100 - 36.27 = 63.73% mu = 149: sigma = 33.3: a = 140: b = 201. Therefore p(a < x < b) = p(140 - 149/33.3 < z < 201 - 149/33.3) = 0.5473 or 54.73 % Therefore required probability is 54.73%