Question

Answer the following question involving "The Normal Distribution and the 68-95-99.7 Rule" and show how I got the answers below.

Answers:

1)a) 68%    b) 47.5%    c) 2.5%

2) 12 or 13 people

Questions:

1) A population of dogs have weights that are normally distributed with an average of 30 pounds with a standard deviation of 3 pounds.

a) What percent of the dogs weigh between 27 and 33 pounds?

b) What percent of the dogs weigh between 30 and 36 pounds?

c) What percent of the dogs weigh less than 24 pounds?

2) In order to qualify for the XYZ Scholarship a student must have a test score that is more than 2 standard deviations above the mean. The mean test score is 73 with a standard deviation of 6. If 500 students take this test, how many will qualify for the XYZ Scholarship?

Answer 1

Given Population of dogs follows normal distribution with standard deviation, mean, u = 30 3 34% 34% 113.547 T13-59X'. na uto M-a M+25 it zo '68% a5.ciul weigh between 27 a) percent of dogs. and 33 pounds. 1 27= u-o 30-3 33 = uito 30 + 3 134% 34% 27. 시 33 30 Answer . 34% + 34% . = 68%

34% + 34 Tit 13.5% of dogs weigh weigh between 27 and b) percent pounds 36 341|3ux 13:57 27 M 33 36 30 alto= 30.+ 3 = 30.+ 3 = 33 ut 20 = 30+283) = 36 u-ï = 30-3 = 27. 47.5% Answer = 24 9 per cent of dogs weight less than pounds: 1-20 = 30-2 (3) = 24 34% 13.51, 34% a 27 30 24 50% Let the requirzed percentage be a

at 13.5% + 34% = 50% x = (50%-47.5%) x= 2.5% 2.5. 2) Normal distribution is followed by Xyz scholarship A student got a test score ; re, which is 2 standard deviations above the mean 73+2 (6) ut.2.0 c = 73+12 > 859 A student need to get above 85 to get qualified for scholarship: 2.5% 348 34% 13.5% TILI Mou M +T. +28 79 85 67 73 The shaded part is required region. The students will be 2.5% of .500 Number of studdent's who gets qualified 12.5 500X 2.5 100 ~ 132 (01.12