Question

Assume that the cans of Coke are filled so that the actual amounts are normally distributed with a mean of 12.00 oz and a standard deviation of 0.11oz. (a) Find the probability that a single can has at least 12.19 g3 (b) Find the probability that a single can has at most 13.00 gz (c) Find the probability that a sample of 26 cans have a mean of at least 12.19 oz (d) If it is required that 60% of the cans filled must pass the standard, what is the cutoff fill level 1. that must be enforced? 2. Based on a survey, 20% of adults believe in reincarnation. Assume that 10 adults are randomly selected, the probability. (a) That exactly five of the selected adults believe in reincarnation (b) At least 7 believe in reincarnation (c) Between 3 and 7 inclusive believe in reincarnation 3. Look at the "Lab Chap 4 Probability Activity 2" and review the solutions. 4. You must be able to identity the sampling techniques that are used, given a scenario: In other words, know the definitions of Stratified, Systematic, Cluster, Random, and Convenience Sampling techniques. You should be able to identify which one is used given a particular scenario. 5. Below is a list of pulse rates from a sample of students: 93 94 68 74 59 88 78 82 86 72 64 72 54 66 56 80 72 6 64 96 59 66 64 84 82 70 64 78 74 86 77 70 96 90 88 85 67 86 82 80 86 59 66 76 59 82 66 75 86 72 64 73 75 72 54 66 64 64 96 70 75 74 69 63 70 Create a frequency distribution whose lower class limit for the first class is 55 and the class width is 5. Determine the mean and standard deviation for the data Describe the shape of the distribution a. b. c. 1 of 3509 Words English (US)

Answer 1

1)

a)

µ = 12       

σ = 0.11

right tailed

X ≥12.190

  

Z =(X - µ ) / σ = (12.19-12) / 0.11=1.72727

  

P(X ≥12.190) = P(Z ≥1.73) =P ( Z <-1.73) = 0.0421(answer)

excel formula for probability from z score is =NORMSDIST(Z)

b)

µ = 12    

σ = 0.11

left tailed

X ≤ 13

  

Z =(X - µ ) / σ =9.09

  

P(X ≤13) = P(Z ≤9.09) =1.0000(answer)

excel formula for probability from z score is =NORMSDIST(Z)

c)

µ = 12       

σ = 0.11

n=26

right tailed

X ≥12.190

  

Z =(X - µ )/(σ/√n) =8.807

  

P(X ≥12.190) = P(Z ≥8.807) =P ( Z <-8.8074) = 0.000000

d)

µ = 12     

σ = 0.11

proportion=0.60

  

Z value at 0.6=0.2533(excel formula =NORMSINV(0.60) )

z=(x-µ)/σ

so, X=zσ+µ=0.253*0.11+12

X=12.03    (answer)