Answer:
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
