1) Processing of the ceramic alloy involves a sintering ( firing ) process which produces a dense hard material. the sintering operation had to be carefully monitored and the tolerance set for the sentering temperature is 1200 c + - 10 c with a standard deviation of 3.6 c. For the investigative work the temperature must be that the probability of the temperature drifting outside tolerance must not exceed 1%.

Using the temperature data supplied assess the temperature capability for the sintering process of trial material.

For this problem i have tried to do the Z score but i do not understand where the mean is .. I think i got the Z which < 0.01 - 1% please help me Do i need to calculate the mean ? please help

Z = X-μ / σ μ- Mean σ Standard deviation

2. In order to launch the new product , special samples kits will be produces . The cost of production will be high initially so it is important not to waste material. However , it is important that the manchines which dispense the modelling and liner liquids do not underfill the containers .

The modelling liquid bottle states a nominal volume of 100 ml . Measeruments you have taken on modelling the liquid volume going into the bottles indicate a mean of 100 and standerd deviation of 0.2 ml. What amount of liquid should be dispensed to limit the risk of underfilling to 1 in 40 ?

For this problem i have calculated The X scores as i got the Z score , than 1/40

0.065 = 0.025 I have checked it on the table of normal destribuition and it falls on between 0.06 and 0.07

1/40 - 0.025 - 2.5 %

P(Z<-0.065)=0.025 P(Z< (X-μ)/σ = 0.025

-0.065= X-100/0.2

-0.065= X/0.2 - 500

-.0065 +500 = x/0.2

499.935 = X/0.2

X =499.935 times 0.2

x =99.987

So for the ascwer of the question 2 its should be dispensed 99.987 ml of the 2.5 % to limit the risk of underfilling

Is the answer correct for this problem .. Please help