Question

7. Suppose that the width of pebbles in a river is normally distributed with mean of 12.1 mm and standard deviation of 3.2 mm. a. What is the probability that X is smaller than 11? b. A random sample of 12 pebbles is taken. What is the probability that X-bar < 11?

Answer 1

Mean width of pebbles ()= 12.1 mm

Mean width of pebbles (mu)= 12.1 mm Standard Deviation of the width of pebbles (sigma) = 3.2 mm a.) Now, Let x = 11mm As our data follow normal distribution curve, the z value at x = 11 is z = x - mu/sigma = 11 - 12.1/3.2 = -0.3438 Now we have to find the P(X < 11mm) which is equal to P(Z < -0.3438) From z table, the probability that P (Z < -0.3438) = 0.3655 b.) Sample size (n) = 12 Now sample standard deviation (s) = sigma/squareroot n = 3.2/squareroot 12 = 0.9238 Now we have to find the probability that x-bar < 11, in that particular sample of 12 pebbles. Now as we have to find the probability in that particular sample, we will take sample standard deviation (s) in all our calculations. Let, x = 11 Now z value will be = z = x - mu/s = 11 - 12.1/0.9238 = -1.1908 (Here we have taken sample standard deviation(s) because we have to find the probability in that particular sample) Now we have to find the P(X < 11mm) which is equal to P(Z < -1.1908) From z table, the probability that P (Z < -1.1908) = 0.1169

Standard Deviation of the width of pebbles ( $\sigma$ ) = 3.2 mm

a.) Now, Let x = 11mm

As our data follow normal distribution curve, the z value at x=11 is

Now we have to find the P(X<11mm) which is equal to P(Z<-0.3438)

From z table, the probability that P (Z<-0.3438) = 0.3655

b.) Sample size (n) = 12

Now sample standard deviation (s) =

Now we have to find the probability that x-bar<11, in that particular sample of 12 pebbles. Now as we have to find the probability in that particular sample, we will take sample standard deviation (s) in all our calculations.

Let, x=11

Now z value will be =

(Here we have taken sample standard deviation(s) because we have to find the probability in that particular sample)

Now we have to find the P(X<11mm) which is equal to P(Z<-1.1908)

From z table, the probability that P (Z<-1.1908) = 0.1169

Please do comment, if you have any doubt.