Question 130143: question 1
According to a survey, 50% of employees in banking sector are satisfied with their jobs. Assume this is the true proportion of all satisfied employees in banking sector. Let p be the proportion of banking sector employees in a random sample of 1000 who are satisfied with their jobs.
a) Describe the shape of the sampling distribution for the sample proportion.
b) Find the mean and standard deviation of the sample proportion, p.
questions 2
Assume that the weights of all mini-packages of a certain brand biscuits are normally distributed with mean of 32 grams and standard deviation of 0.3 grams. Find the probability that the mean weightx, of a random sample of 20 packages of this brand of biscuits will be between 31.8 and 31.9 grams.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! According to a survey, 50% of employees in banking sector are satisfied with their jobs. Assume this is the true proportion of all satisfied employees in banking sector. Let p be the proportion of banking sector employees in a random sample of 1000 who are satisfied with their jobs.
a) Describe the shape of the sampling distribution for the sample proportion.
The sample proportions are normally distributed.
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b) Find the mean and standard deviation of the sample proportion, p.
The mean is 50% or 1/2; The standard deviation is sqrt{pq/n]
= sqrt[(1/2)^2/1000] = [(1/2)/10]sqrt(1/10) = (1/20)[sqrt(10)/10]
= (1/200)sqrt(10)
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questions 2
Assume that the weights of all mini-packages of a certain brand biscuits are normally distributed with mean of 32 grams and standard deviation of 0.3 grams. Find the probability that the mean weightx, of a random sample of 20 packages of this brand of biscuits will be between 31.8 and 31.9 grams.
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Find the z-score of 31.8 and 31.9
z(31.8) = (31.8-32)/[0.3/sqrt(20)] = -0.2*sqrt(20)/0.3 = -2.9814
z(31.9) = (31.9-32)/[0.3/sqrt(20)] = -0.1*sqrt(20)/0.3 = -1.490712
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P(31.8 < x < 31.9) = P(-2.9814 < z < -1.4907) = 0.06658...
Cheers,
Stan H.
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