The y intercept is where m=0. Hence, set m=0 and solve for c:
c=0.05(0)+4.95
c=4.95
It y=4z-3/6 BOOMM done for ya
Answer:
68% of pregnancies last between 250 and 282 days
Step-by-step explanation:
The Empirical Rule states that, for a normally distributed random variable:
68% of the measures are within 1 standard deviation of the mean.
95% of the measures are within 2 standard deviation of the mean.
99.7% of the measures are within 3 standard deviations of the mean.
In this problem, we have that:
Mean = 266
Standard deviation = 16
What percentage of pregnancies last between 250 and 282 days?
250 = 266 - 16
250 is one standard deviation below the mean
282 = 266 + 16
282 is one standard deviation above the mean
By the Empirical Rule, 68% of pregnancies last between 250 and 282 days
Answer:
Theoretically, since it has two sides, it's a 50 50, chance. 50%
A. The point estimate would be the average of the interval boundaries, which is the average of 48.2 and 56.4. This gives a point estimate of 52.3%.
b. The margin of error is the distance from either interval boundary to the point estimate. 56.4 - 52.3 = 4.1%.
c. Assuming this is solely going to be based on the class' decision, their grades should be scored on a curve. Although the margin of error is high and this seems to not have a significant distance from 50%, there is no "middle ground" in this option. We must either grade on a curve or not, and if we must choose one, we have to use to point estimate that is just slightly above 50%.
