Question

Question 2 The cooling oil of a particular machine in a factory is effective (useable) according to a normal distribution with mean of 50 days and standard deviation of 16 days. a) What is the probability that the oil is still effective after 60 days? b) What is the probability that the oil is effective less than 40 days? c) The factory policy is to change the oil at a time (in days) that the probability that the oil is still effective after that day is 0.3. Calculate after how many days the factory should change the oil? d) What is the probability that the oil is still effective after 60 days, if we know that it was effective more than 40 days?

Answer 1

Let X be the random variable corresponding to the number of days the oil remains effective

X is normally distributed with mean 50 and standard deviation 16 days

Probability in normal distribution can be calculate by finding the Z score

Z = (X - Mean) / Std deviation

Use a normal distribution table or a calculator to find the corresponding probabilities

A) P(X > 60) = P( Z > (60-50)/16 ) = P( Z> 0.625 ) = 0.266

B) P(X<40) = P(Z < (40-50)/16) = P( Z < -0.625) = 0.266

C) Oil will be changed after on the day after which probability of remaining effective is 0.3.

Let it be x days

P( X > x ) = 0.3

P( Z > (x -50) / 16 ) =0.3

Z from the probability table comes out to be 0.524

(x - 50) / 16 = 0.524

x = 58.38 days

round it to 58 days

D) Probability of remaining effective after 60 days given it was already effective for 40 days

P( X>60 | X >40 ) = P( X > 60 $\tiny \bigcap$ X > 40 ) / P( X >40 )

Intersection region 40 50 60

P( X > 60 $\tiny \bigcap$ X > 40 ) is the intersection region

Intersection region : P(X > 60 ) = P( Z > (60-50)/16) = P( Z > 0.625) = 0.266

P ( X > 40) = P( Z > (40-50)/16) = P( Z > -0.625) = 0.734

Therefore, P( X>60 | X >40 ) = 0.266/0.734 = 0.362