Question

The manager of a cultured pearl farm has received a special order for five pearls between 7 millimeters and 9 millimeters in diameter. from past experience, the manager knows that the pearls found in his oyster bed have diameters that are normally distributed with a mean of 8 millimeters and a standard deviation of 2.2 millimeters. assume that every oyster contains one pearl. what is the probability that the special order can be filled after randomly selecting and opening the first five oysters?

Answer 1

Answer:

Probability is 0.3506

Step-by-step explanation:

We are given

mean is 8mm

so, $\mu=8$

standard deviation is 2.2 mm

so, $\sigma=2.2$

x-varies from 7mm to 9mm

At x=7:

we can use formula

$z=\frac{x-\mu}{\sigma}$

we can plug values

$z=\frac{7-8}{2.2}$

$z=-0.45455$

At x=9:

we can use formula

$z=\frac{x-\mu}{\sigma}$

we can plug values

$z=\frac{9-8}{2.2}$

$z=0.45455$

So, z is between -0.45455 and 0.45455

$-0.45455\leq z \leq 0.45455$

now, we can use normal distribution table

and we get

Probability is 0.3506