Jason took his Siberian Husky dog to the veterinarian. The dog was put on a scale and found to weigh 52 lb. Assume that the population mean weight of Siberian Huskies is 49 lb. with a standard deviation of 7. The weights of Huskies are normally distributed.
What is the approximate percentage of Huskies that weigh less than Jason's dog?
a) 67%
b) 40%
c) 33%
d) 60%
the weight is less than half of a s.d. above the mean
a) looks good
yikes meant (A) lol
okay, Ima take the test and see if the answer R_scott was correct.
Grammar problems yikes
D) 67%
To find the approximate percentage of Huskies that weigh less than Jason's dog, we need to calculate the z-score for Jason's dog's weight and then use the z-score to find the corresponding percentage.
The z-score is calculated using the formula: z = (X - μ) / σ, where X is the value we are interested in (Jason's dog's weight), μ is the population mean weight of 49 lb, and σ is the standard deviation of 7.
In this case, X = 52 lb, μ = 49 lb, and σ = 7.
Calculating the z-score:
z = (52 - 49) / 7
z = 3 / 7
z ≈ 0.43
Next, we need to find the corresponding percentage using the z-score. We can use a standard normal distribution table or a calculator to do this.
Looking up the z-score of 0.43 in a standard normal distribution table (or using a calculator), we find that the percentage associated with this z-score is approximately 66.76%.
Therefore, the approximate percentage of Huskies that weigh less than Jason's dog is approximately 66.76%.
Since this is closest to 67%, the correct answer is option a) 67%.