Problem 209SE

Without an automated irrigation system, the height of plants two weeks after germination is normally distributed with a mean of 2.5 centimeters and a standard deviation of 0.5 centimeter.

(a) What is the probability that a plant’s height is greater than 2.25 centimeters?

(b) What is the probability that a plant’s height is between 2.0 and 3.0 centimeters?

(c) What height is exceeded by 90% of the plants?

Solution 209SE

Step1 of 4:

Let us consider a random variable X presents the the height of plants two weeks after germination. And X is normally distributed with mean and standard deviation

Here our goal is:

a). We need to find the probability that a plant’s height is greater than 2.25 centimeters.

b). We need to find the probability that a plant’s height is between 2.0 and 3.0 centimeters.

c). We need to find the height which is exceeded by 90% of the plants.

Step2 of 4:

a).

Consider,

Where, is obtained from standard normal table(area under normal curve).

(In area under normal curve we have to see in row -0.5 under column 0.00)

Hence,

Therefore,

Step3 of 4:

b).

Consider,

Where, is obtained from standard normal table(area under normal curve).

(In area under normal curve we have to see in row -1.0 under column 0.00)

(In area under normal curve we have to see in row 1.0 under column 0.00)

Hence,

Therefore,

Step4 of 4:

c).

We have (1 - ) = 0.90,

= 1 - 0.90

= 0.10

Hence,

Where, is obtained from standard normal table(area under normal curve). In area under normal curve we have see where 0.10 value falls, it falls in row -1.2 under column 0.08.

The corresponding value is the mean increased by the product of the Z score and the standard deviation:

Therefore, x = 1.86.