Let be the sample proportion of success in a random sample from a population with proportion of success p.

Then

1. The sampling distribution of has mean p

.

2. The sampling distribution of has a standard deviation

if

if .

3.The value for a value of is

.

Once again, the last approximation is good if .

Example:

The firm makes deliveries of a large number of products to its customers.

It is known that 75% of all the orders it receives from its customers are delivered on time. Let be the proportion of orders in a random sample of 120 that are delivered on time. Find the probability that the value of Summary. Let be the sample proportion of success in a random sample from a population with proportion of success p. will be

a) between 0.73 and 0.80;

b) less than 0.72.

Solution:

From the given information,

, ,

where p is the proportion of orders in the population.

The mean of the sample proportion is

The standard deviation of is

.

Let us find .

.

Since , we can infer from the central limit theorem that the sampling distribution of is approximately normal.

Next, the two values of are converted to their respective Z vales by

.

a) For ; .

For ; .

The required probability is (Figure 5.5).

.

Thus, the probability is 0.7011 that between 73% and 80% of orders of the sample of 210 orders will be delivered on time.

b)

.

Thus, the probability that less than Summary. Let be the sample proportion of success in a random sample from a population with proportion of success p. 72% of the sample of 210 orders will be delivered on time is 0.1567.

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Документ Summary. Let be the sample proportion of success in a random sample from a population with proportion of success p.