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.
So we solve for each
70% of 100=70
students who didn't buy lunch=100-70=30
some of the 70 left so that the students who bought lunch, are 60% of those who are left
we know that the students who didn't buy lunch didn't change number so therefor
100% of students left-60%=40%= students that didn't buy lunch=30 so
40%=30
20%=15
100%=75
there were 75 students lfet in the cafeteria
25.69 + .75= $26.44
Hope this helps.
That would be week 2 and 4 (B). They are both at 12 students attending
Hope this helped!