5. Group of people aged 15-65 are randomly selected and arranged in group of six

Question

5. Group of people aged 15-65 are randomly selected and arranged in group of six. The random variable x is the number in the group who say that their family contribute most to their happiness. (a) Is this a probability distribution? (b) If "yes" find the mean and standard deviation. (c) If"not" give the reason. f(x) 0+ 0.003 0.025 0.279 0.373 0.208 4 6. Mars, Inc. claims that 14% of its M&M; plain candies are yellow, and a sample of 100 such candies is randomly selected. Assume that a procedure yields a binomial distribution, so n-100 and p-0.14. (a) Find the mean and standard deviation for the number of yellow candies. Mean n Diancard Desin (b) What does Chebyshev's theorem with k candies. 2 tell us about the number of yellow

Explanation / Answer

Ans:

5)

a)Yes,it is probabity distribution,as,sum of all probabilities is 1.

b)

n=6,p=0.77

we can also use, mean=np=4.62

standard deviation=sqrt(np(1-p))=1.03

6)mean=np=14

standard deviation=3.5

b)when k=2

1-(1/2^2)=1-(1/4)=0.75 i.e. 75% of the data lies within 2 standard deviations of the mean.

14-2*3.5=7

14+2*3.5=21

so,75% of the data lies within 7 and 21.

x p(x) x*p(x) (x-4.62)^2*p(x) 0 0 0 0.000 1 0.003 0.003 0.039 2 0.025 0.05 0.172 3 0.111 0.333 0.291 4 0.279 1.116 0.107 5 0.374 1.87 0.054 6 0.208 1.248 0.396 Total 1 4.62 1.0596 mean variance standard dev.= 1.03

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