Answer:
-1.683
Step-by-step explanation:
Given :
Group 1 :
x1 = 667 ; n1 = 10 ; s1 = 20
Group 2 :
x2 = 679 ; n2 = 14 ; s2 = 15
The test statistic assuming equal variance :
x1 - x2 / √[Sp² * (1/n1 + 1/n2)]
sp² = [(n1 - 1)*s1² + (n2 - 1)*s2²] ÷ (n1 + n2 - 2)
Sp² = [(10 - 1)*20² + (14 - 1)*15²] = 296.59
Test statistic =
(667 - 679)/ √[296.59 * (1/10 + 1/14)]
-12 / 7.1304978
Test statistic = - 1.682
(1/3) cubed, since the equation for the volume of a cube is an edge, cubed
Answer:
68.26% probability that a randomly selected full-term pregnancy baby's birth weight is between 6.4 and 8.6 pounds
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:

What is the probability that a randomly selected full-term pregnancy baby's birth weight is between 6.4 and 8.6 pounds
This is the pvalue of Z when X = 8.6 subtracted by the pvalue of Z when X = 6.4. So
X = 8.6



has a pvalue of 0.8413
X = 6.4



has a pvalue of 0.1587
0.8413 - 0.1587 = 0.6826
68.26% probability that a randomly selected full-term pregnancy baby's birth weight is between 6.4 and 8.6 pounds
Answer: 14
Step-by-step explanation: 9 small squares with one teddy bear in each, 4 medium squares with 4 bears in each medium square, and 1 big square for the entire thing.