Question

Dartmouth Distribution Warehouse makes deliveries of a large number of products to its customers. It is known that 81% of all the orders it receives from its customers are delivered on time. Let p^ 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 between 0.78 and 0.86.

Round your answer to four decimal places.

Answer 1

Solution

Given that,

$\sigma$
_р
=  $\sqrt$[p ( 1 - p ) / n] = $\sqrt$ [(0.81 * 0.19) / 100 ] = 0.0392

= P[(0.78 - 0.81) /0.0392 < (
р
- $\mu$
_р
) / $\sigma$
_р
< (0.86 - 0.81) / 0.0392]

= P(-0.77 < z < 1.28)

= P(z < 1.28) - P(z < -0.77)

= 0.8997 - 0.2206

= 0.6791

Probability = 0.6791