Suppose that large dogs weights are normally distributed with a mean of 100 pounds and a standard deviation of 10 pounds. If one of these dogs is picked random:

Question

Suppose that large dogs weights are normally distributed with a mean of 100 pounds and a standard deviation of 10 pounds. If one of these dogs is picked random:

a) what is the probability that the dog will weigh over 110 pounds?
b) what is the probability that the dog will weigh less than 90 pounds?
c) what is the probability that the dog's weight will be between 90 and 100 pounds?
d) What is the probability that the dog's weight will be between 90 and 110 pounds?
e) If a random sample of 36 dogs is taken rom this population, will the sampling distribution of the mean of that sample have a normal distribtion

Answer 1

I love this little applet for these kind of questions

http://davidmlane.com/hyperstat/z_table.html

I really see no difference using this as opposed to looking up values from a table.

Answer 2

To answer these questions, we need to use the concept of the normal distribution and Z-scores. The Z-score measures the number of standard deviations a particular value is from the mean.

To calculate probabilities for normally distributed variables, we can use the Z-table or a statistical calculator.

Let's now calculate the probabilities step by step:

a) Probability that the dog will weigh over 110 pounds:
First, we calculate the Z-score:
Z = (110 - 100) / 10 = 1

Using the Z-table, we find the area to the right of Z = 1. This corresponds to the probability of the dog weighing over 110 pounds.

b) Probability that the dog will weigh less than 90 pounds:
Similarly, we calculate the Z-score:
Z = (90 - 100) / 10 = -1

Using the Z-table, we find the area to the left of Z = -1. This corresponds to the probability of the dog weighing less than 90 pounds.

c) Probability that the dog's weight will be between 90 and 100 pounds:
To find this probability, we need to find the area between two Z-scores.

First, we calculate the Z-scores for 90 pounds and 100 pounds:
Z1 = (90 - 100) / 10 = -1
Z2 = (100 - 100) / 10 = 0

Using the Z-table, we find the area to the left of Z = 0 and subtract the area to the left of Z = -1. This gives us the probability of the dog's weight being between 90 and 100 pounds.

d) Probability that the dog's weight will be between 90 and 110 pounds:
Similar to the previous question, we calculate the Z-scores:
Z1 = (90 - 100) / 10 = -1
Z2 = (110 - 100) / 10 = 1

Using the Z-table, we find the area to the left of Z = 1 and subtract the area to the left of Z = -1. This gives us the probability of the dog's weight being between 90 and 110 pounds.

e) If a random sample of 36 dogs is taken from this population, will the sampling distribution of the mean of that sample have a normal distribution?
Yes, according to the Central Limit Theorem, as long as the sample size is sufficiently large (typically n > 30) and the population from which the sample is drawn is not highly skewed, the sampling distribution of the mean is approximately normally distributed. In this case, since the sample size is 36, we can assume the sampling distribution of the mean will be approximately normally distributed.