Could anybody help me with this i will cashapp you if you get it right
1 answer:
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Complete question:
Assume that different groups of couples use a particular method of gender selection and each couple gives birth to one baby this mention is designed to increase the likelihood that each baby will be a girl, but assume that the method has no effect so the probability of a girl is 0.5. Assume that the group consists of 36 couples.
A) Find the mean and standard deviation for the number of girls in groups of 36 births.
B) Use the range rule of thumb to find the values separating results that are significantly low and significantly high.
C) Is the result of 33 girls significantly high? A result of 33 girls would suggest the method is effective or is not effective?
Answer:
a) mean = 18
Standard deviation =3
b) low range = 12
High range = 24
c) The result of 33 girls is significantly high. Yes, the method is effective.
Step-by-step explanation:
Given:
p = 0.5
n = 36
a) The mean is the product of n and p
Mean u = np
u = 36 * 0.5 = 18
The standard deviation is the square root of the product of n and p&q.
S.d ó =
b) To find the range rule of thumb:
• For low range
Low range = u - 2ó
= 18 - (2 * 3)
= 12
• High range
= u + 2ó
= 18 + (2*3)
= 24
c) The result is significantly high, because 33 is greater than 24 girls.
A result of 33 girls would prove the method as effective.
Answer:
The following are the answer to this question:
Step-by-step explanation:
In the question some data is missing, which is defined in the attached file please find it.
The table for point a:
plot for the Stem-and-leaf:
0 3 4 5 7 4
1 2 2 4 4 5 7 7 7
2 0 3 4 4 5 8 6
3 3 4 7 9 4
4 1
In point b:
Its distributed skewed is correct because in its points 1 is a tail on its right side.
please find the attached file.
