Question

myOpenMath Home | My Classes - User Settings Log Out Course Messages Forums Calendar Gradebook Home > MATH-140 Statistics U20 > Assessment 4.3 Practice Problems Score: 17/36 2/5 answered DC Question 4 < > B0/8 pts The percent of fat calories that a person in America consumes each day is normally distributed with a mean of about 30 and a standard deviation of about 7. 1. Suppose one individual is randomly chosen. Find the probability that the individual normally consumes between 28 and 32 fat calories per day. . What distribution will you use to calculate this probability? NG o List the z-scores needed to calculate the result. If there is more than one z-score, separate the values with a comma. • There is a 28 and 32 calories per day. chance that the individual normally consumes between 2. Suppose that 26 individuals are randomly chosen. Find the probability that they normally consume an average of between 28 and 32 calories per day. . What distribution will you use to calculate this probability? NO o List the z-scores needed to calculate the result. If there is more than one-score, separate the values with a comma, There is a chance that the 26 individuals normally consume an average of between 28 and 32 calories per day Question Help: D Post to forum Submit Question

I'm confuse how to get 1 and 2 as I struggled with the past queations

Answer 1

Moan of fat calories consuered per day r= 30 standard deviation o=7 The Normal Distribution N(H, o) is used to calculate the probability and the distribution is N( 30, 7). Answer Au sidividual Nonwally consumes between 28 and 39 fat calories per day for x = 28, Z-score 28-30 7 -0.2857 for x = 32, X-4 32-30 7 = + 2 / = + 0.2857 Z-score =(0.2857 is needed to calculate the result. Auswes X-M - 24 Z 2. score = z Thus P (28 <<<32) P 402851(2<+0.2857) = 0.22489 = 22.489% Prehability that an individual consumes fat calories between 28 cal, and 32 cal.is Answer 22.489%

2. no 26 11 / 1 Population mean M = 30 Sample size standard Deviation o=7 Sanuple standard deviation o a Distribution used to calculate probability, N~(M, NN (30, 20) Answer 28 < X < 32 for X=28 メーM Z. Scare = 28-30 28-30 (7126) 5.1 Z 5--0.39 z. score Z for X=30 メーM 32-30 32-30 ( on) (7/126) 5.1 = + 2 + 0.39 5.1 is needed to calculate the Answer Thus, 2-score 0.39 probability P (28<x<32) P (-0.39<<<+0.39) 0.30346 30.346% Thus Probability that an individual consumes fat calosies 30.346% between 30 & 32 is Auswer