Question

Dartmouth Distribution Warehouse makes deliveries of a large number of products to its customers. It is known that 78% of all the orders it receives from its customers are delivered on time. Let o be the proportion of orders in a random sample of 100 that are delivered on time. Find the probability that the value of P will be less than 0.81

Answer 1

The probability that the value of P will be less than 0.81 is 0.7642.

Explanation:

According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of this sampling distribution of sample proportion is:

`\mu_(\hat p)=p`

The standard deviation of this sampling distribution of sample proportion is:

`\sigma_(\hat p)=\sqrt{(p(1-p))/(n)}`

Given:

n = 100

p = 0.78

Since n = 100 > 30, according to the central limit theorem the sampling distribution of sample proportion follows a Normal distribution.

`\hat p\sim N(\mu_(\hat p)=0.78, \sigma_(\hat p)=0.0414)`

Compute the value of
`P(\hat p<0.81)` as follows:

`P(\hat p<0.81)=P((\hat p-\mu_(\hat p))/(\sigma_(\hat p))<(0.81-0.78)/(0.0414))`

`=P(Z<0.72)\\=0.76424\\\approx0.7642`

*Use a z-table for the probability.

Thus, the probability that the value of P will be less than 0.81 is 0.7642.