Question

The Australian Bureau of Statistics has reported that in 2015, 26.78% of the 16.8 million employees in Australia worked part-time in their main job. Suppose that you select a random sample of 250 employees from around Australia.

(a) What is the probability that more than 26.2% of those sampled work part-time in their main job?

(b) What is the probability that the proportion of part-time employees in the sample is between 27% and 29%?

(c) The probability is 77% that the sample proportion of part-time workers will be above what value?

Answer 1

1)

Population Proportion =	0.2678
Sample Size =	250

We need to compute  .

the population men of sample proportions and the corresponding standard error are:

  

Observe that:

which indicates that the assumption for normal approximation for the sampling distribution is met.

Now, the following is obtained using normal approximation:

2)

We need to compute  

the population men of sample proportions and the corresponding standard error are:

Observe that:

which indicates that the assumption for normal approximation for the sampling distribution is met.

Now, the following is obtained using normal approximation:

3)

We need a point above which probability is 77% i.e 0.77.

Mathematically, Pr (X > x) = 0.77 which can be written as Pr(X < x) = 0.23

the population men of sample proportions and the corresponding standard error are:

Observe that:

np  = 250⋅0.2678 = 66.95 ≥ 10 ,   nq= 250  ⋅  0.7322 = 183.05 ≥ 10

which indicates that the assumption for normal approximation for the sampling distribution is met.

We need to find a score x so that the corresponding cumulative normal probability is equal to 0.23. Mathematically, x is such that:

3Pr(X≤x)=0.23

The corresponding z score score so that the cumulative standard normal probability distribution is 0.23 is

zc​ = - 0.7388

This value of zc​ = -0.7388 can be found either with Excel, or with a normal distribution table. Hence, the X score associated with the 0.23 cumulative probability is

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