Question

The randomness in measured values of the thermal conductivity of the material is treated as a normal random variable with mean 29.567 W/( mK) and a standard deviation of 1.507 W/(mK). (a) What is the probability of thermal conductivity having a value between 28 W/(m K) and 29 W/(mK)? (b) What is the value of thermal conductivity with a 25% probability?

Answer 1

(a)

$\mu$ =29.567

$\sigma$ = 1.507

To find P(28 < X < 29):

For X = 28:

Z = (28 - 29.567)/1.507

= - 1.0398
By Technology, Cumulative Area Under Standard Normal Curve = 0.1492

For X = 29:

Z = (29 - 29.567)/1.507

= - 0.3762
By Technology, Cumulative Area Under Standard Normal Curve = 0.3534

So,

P(28 < X < 29):= 0.3534 - 0.1492 = 0.2042

So,

Answer is:

0.2042

(b)

Value of thermal conductivity with a 25% probability corresponds to Cumulative Area Under Standard Normal Curve= 0.25.

By Technology, corresponding Z score = - 0.6745

So,

we have:

Z = - 0.6745 = (X - 29.567)/1.507

So,

X = 29.567 - (0.6745 X 1.507)

= 28.5505

So,

Answer is:

28.5505