Question

Gobblers, a small midwestern clothier, is considering fabricating a new line of men's pants. It needs some help understanding its current male customer base in order to better assess what sizes of pants to offer. The firm presently has demographic data on all 5,000 of its male customers. It has found that the average weight of all its male customers is 195 lbs with a standard deviation of 20 pounds. In addition, the average weight has been found to be normally distributed. Listed below you will find weight information on just 20 randomly sampled customers. Customer Weight 1 200 lbs 2 225 3 180 4 197 5 186 6 239 7 167 8 143 9 242 10 194 11 201 12 185 13 184 14 178 15 181 16 210 17 203 18 194 19 212 20 207
5.) Based upon the population data, what percentage of customers would be expected to weigh above 230 pounds? 6.)Basd upon the population data what is the weight value that you would expect 25 percent of the customers to exceed? 7.) Based upon the population data what is the weight level that you would expect 90% of the customers to exceed? 8.)based upon the population data what percentage of customers would be expected to weigh between 170 and 230 lbs?

Answer 1

5) P(X > 230)

    = P((X - )/ > (230 - )/)

    = P(Z > (230 - 195)/20)

    = P(Z > 1.75)

    = 1 - P(Z < 1.75)

    = 1 - 0.9599

    = 0.0401

6) P(X > x) = 0.25

   or, P((X - )/ > (x - )/) = 0.25

   or, P(Z > (x - 195)/20) = 0.25

   or, P(Z < (x - 195)/20) = 0.75

   or, (x - 195)/20 = 0.67

   or, x = 0.67 * 20 + 195

   or, x = 208.4

7) P(X > x) = 0.9

   or, P((X - )/ > (x - )/) = 0.9

   or, P(Z > (x - 195)/20) = 0.9

   or, P(Z < (x - 195)/20) = 0.1

   or, (x - 195)/20 = -1.28

   or, x = -1.28 * 20 + 195

   or, x = 169.4

8) P(170 < X < 230)

    = P((170 - )/ < (X - )/ < (230 - )/)

    = P((170 - 195)/20 < Z < (230 - 195)/20)

    = P(-1.25 < Z < 1.75)

    = P(Z < 1.75) - P(Z < -1.25)

    = 0.9599 - 0.1056

    = 0.8543