Question 130143
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.
----------------
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)
------------------------- 
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.
----------
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
----------------
P(31.8 < x < 31.9) = P(-2.9814 < z < -1.4907) = 0.06658...
Cheers,
Stan H.