According to a website, the mean weight of an adult member of a particular breed of dog is 145 pounds. Assume the distribution of weights is normal with a standard deviation of 12 pounds.

a. Find the standard score associated with a weight of 169 pounds. (Type an integer or decimal rounded to two decimal places as needed.)

b. Using the empirical rule, what is the probability that a randomly selected dog of this breed weighs more than 169 pounds? Use technology to confirm your answer is correct. (Type an integer or decimal rounded to one decimal place as needed.)

c. Almost all adult dogs of this breed will have weights between what two values?

The standard score of a dog weight of 169 pounds is 2. The probability of a dog weighing more than 169 pounds is 2.28%. Almost all dogs of this breed will weight between 109 and 181 pounds.

Explanation:

The Normal Distributions, Standard Scores (also known as Z-Scores), and Probabilities. Given a mean weight of 145 pounds and a standard deviation of 12 pounds, the Z-Score for a weight of 169 pounds can be found using the Z-Score formula: Z = (X - μ) / σ where X is the weight, μ is the mean, and σ is the standard deviation. So, Z = (169 - 145) / 12 = 2.0.

Using Z-Score table or a technology tool (like a calculator with normal distribution functions), we find the probability that a randomly selected dog of this breed weighs more than 169 pounds is the area to the right of our Z-Score, which is about 0.0228 or 2.28%.

Finally, because almost all values in a Normal Distribution fall within 3 standard deviations of the mean, the weights of nearly all dogs in this breed will be between 145 - 3(12) = 109 pounds and 145 + 3(12) = 181 pounds.

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