1. A statistician calculates that 8% of Americans are vegetarians. If the statistician is correct, what is the probability that the proportion of vegetarians in a sample of 569 Americans would be greater than 9%?2.The mean points obtained in an aptitude examination is 166 points with a standard deviation of 19 points. What is the probability that the mean of the sample would differ from the population mean by less than 1.1 points if 66 exams are sampled?

1) the probability that the proportion of vegetarians in the sample would be greater than 9% is 0.1902

2) the probability that the mean of the sample would differ from the population mean by less than 1.1 points in the is 0.3616

p( /x" - u/ < 0.47 ) = 0.3616

Step-by-step explanation:

1)

Given that;

p = 8% = 0.08 and n = 569

what is the probability that the proportion of vegetarians in a sample of 569 Americans would be greater than 9% (0.09)

that is;

p( p" > 0.09) = ?

p( p" > 0.09) = p( z > [(0.09 - 0.08)/(√((0.08(1 - 0.08))/569))] )

= p( z > (0.01 / 0.0114))

= p ( z > 0.8772)

= 0.1902

Therefore the probability that the proportion of vegetarians in the sample would be greater than 9% is 0.1902

2)

Given that;

mean points obtained in an aptitude examination u = 166

standard deviation α = 19

sample size n = 66

x" - u = 1.1

first we say

x-score = (x" - u) / (α/√n) = 1.1 / (19/√66) = 1.1 / 2.3387 = 0.47

p( -0.47 < z < 0.47 )

= p( z < 0.47) - p( z < - 0.47)

= 0.6808 - 0.3192

= 0.3616

Therefore the probability that the mean of the sample would differ from the population mean by less than 1.1 points in the is 0.3616

p( /x" - u/ < 0.47 ) = 0.3616

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