Final answer:
A score of 74 has a z-score of 0.3333 for a normal distribution with mean 70 and standard deviation 12. The body to the left of this score contains approximately 62.93% of the distribution.
Step-by-step explanation:
To determine whether the body of a normal distribution is to the left or right of a score, and to find the proportion of the distribution in that body, we first calculate the z-score. The z-score tells us how many standard deviations away from the mean a raw score is. For a normal distribution with a mean (μ) of 70 and a standard deviation (σ) of 12, a score (X) of 74 would have a z-score (z) calculated as follows: z = (X - μ) / σ = (74 - 70) / 12 = 4 / 12 = 0.3333.
Since 74 is above the mean of 70, the body of the distribution is to the left of the score. The proportion of the distribution in the body to the left of the z-score of 0.3333 can be found using a standard normal distribution table or software. This z-score corresponds to an area of approximately 0.6293, which means that approximately 62.93% of the distribution lies to the left of the score of 74.