Answer to Question #254585 in Statistics and Probability for Maizatul

Question

two independent experiments are being conducted in which two different types of drying methods, freeze drying and oven drying are compared. 18 fruits are dried using the freeze drying and the drying time is recorded. the same procedure is done with oven drying. the standard deviation for both methods are known to be 30 minutes. Assume that the mean drying time for both methods are the same and the time taken for fruits to dry using freeze drying method and oven drying method are normally distributed. Find the probability that

a) the sample mean for freeze drying is less than the sample mean for oven drying.

b)the difference between both sample means is larger than 2 minutes.

c)the difference between both sample means is less than1 minutes.

Answer 1

Let $u{\scriptscriptstyle 1},u{\scriptscriptstyle 2}$ - sample means of freeze and oven samples respectively.

$u{\scriptscriptstyle 1} = {\frac {n{\scriptscriptstyle 1}+...+n{\scriptscriptstyle 18}} {18}}$ , where n - values of the sample elements

Since n ~ $N(x, 30)$, then $u{\scriptscriptstyle 1}$ ~ N(x, ${\frac {18*30^{2}} {{18^{2}}}}$) = N(x, 50), $u{\scriptscriptstyle 2}$ has the same distribution

Then $u{\scriptscriptstyle 1}-u{\scriptscriptstyle 2}$ ~ N(0, 100) = 10N(0,1)

a) The sample mean for freeze drying is less than the sample mean for oven drying

P(10N(0,1) < 0) = 0.5

b) P(-10N(0,1) < -2 and 10N(0,1) > 2) = P(N(0,1)< -0.2) + P(N(0,1) > 0.2) = 2P(N(0,1) < -0.2) = 0.84148

c) P(-1 <10N(0,1) < 1) = P(-0.1 < N(0,1) < 0.1) = P(N(0,1) < 0.1) - P(N(0,1) < -0.1) = 0.53983 - 0.46017= 0.07966