The world\'s smallest mammal is the Kitti\'s hog-noised bat, with a mean weight

Question

The world's smallest mammal is the Kitti's hog-noised bat, with a mean weight of 1.5 grams and a standard deviation of 0.25 grams. Assuming that the weights are normally distributed, find the probability for randomly selecting a bat that weighs between 1.0 and 2.0 grams. (Notice we are taking a sample size of one!)

Input 1.5 for the mean and 0.25 for the std. dev. Then make sure the pulldown option for P(x), or Prob(x), is set for '<= 2' (You can set the pulldown to '<=' to get the area to the left and '=>' to get the area to the right). Click on 'Compute'. You should get an answer of '0.97724...'.  Then go back and set it for <= 1. You should get an answer of '0.02275...'.   If you subtract these two values you will get the P(1
z = (1-1.5)/.25 = -2 and z = (2-1.5)/.25 = +2

Notice the normal calculator is finding the area between z = 2 and z = -2 just as you would do in the examples in the text and section exercises. When you use the normal calculator, it is just giving the area you want under the normal curve.

Now let us consider that we take a random sample of 7 of Kitti's hog-nosed bats.

3. Now find probability that the sample mean falls between 1.6 and 2.2 grams.   What value will you now use for the standard deviation?  (Find the Standard Error of the Mean!) Explain and show what you did to find it. After each computation with the normal calculator, Click on Ctrl, Alt, Print Screen to capture your screen and paste it into your solution file. Then show your steps in finding the final answer from your data.

4. Find probability that the sample mean is more than 2.2 grams.   What value will you again use for the standard deviation? (Find the Standard Error of the Mean!) After each computation with the normal calculator, Click on Ctrl, Alt, Print Screen to capture your screen and paste it into your solution file. Then explain how you got the solution based on the results from the normal calculator.

Explanation / Answer

3) Se of mean = 0.25 / srqt 7 = 0.0945

P( 1.6 - 1.5 / 0.0945 < z < 2.2 - 1.5 / 0.0945 )

P( 1.06 < z < 7.41 ) = 0.1446 - 0.0000 = 0.1446

4)

P( z > 2.2 - 1.5 / 0.0945 )

P( z > 7.41) = 0.0000

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