Help me with this one please and thank you
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Answer:
The 99% confidence interval estimate of the percentage of girls born is (74.37%, 85.63%).
Usually, 50% of the babies are girls. This confidence interval gives values considerably higher than that, so the method to increase the probability of conceiving a girl appears to be very effective.
Step-by-step explanation:
Confidence Interval for the proportion:
In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of
, we have the following confidence interval of proportions.
In which
z is the zscore that has a pvalue of .
For this problem, we have that:
99% confidence level
So , z is the value of Z that has a pvalue of
, so
.
The lower limit of this interval is:
The upper limit of this interval is:
For the percentage:
Multiplying the proportions by 100.
The 99% confidence interval estimate of the percentage of girls born is (74.37%, 85.63%).
Usually, 50% of the babies are girls. This confidence interval gives values considerably higher than that, so the method to increase the probability of conceiving a girl appears to be very effective.
Answer: p >
Step-by-step explanation:
-21p + 69 > -49p + 81
-69 -69
-21p > -49p + 12
+49p +49p
28p > 12 divide both sides by 28
p > 12/28
reduce it p> 3/7