The probability of a female birth is about 0.5. over the period of a year, would there be more days when at least 60% of the babies born were girls in
a.a large hospital
b.a small hospital
c. it makes no difference
please explain your reasoning
thanks
To determine whether there would be more days with at least 60% of babies being girls in a large hospital, small hospital, or if it makes no difference, we need to calculate the probabilities for each scenario.
Let's start by assuming that the probability of a female birth remains constant throughout the year at 0.5. We can use the binomial distribution to calculate the probability of having a certain number of girls out of a total number of births.
a) Large Hospital:
In a large hospital, the total number of births in a year is likely to be significantly higher compared to a small hospital. Let's assume the large hospital has 1,000 births per year. To calculate the probability of having at least 60% of babies being girls, we need to calculate the probability of having 600 or more girls out of 1,000 births.
Using the binomial distribution formula, we can calculate this probability as a sum of individual probabilities:
P(at least 600 girls) = P(600 girls) + P(601 girls) + ... + P(1000 girls)
P(X=k) = (nCk) * p^k * (1-p)^(n-k)
where n is the total number of births, k is the number of girls, nCk is the binomial coefficient (n choose k), p is the probability of a female birth (0.5 in this case), and (1-p) is the probability of a male birth.
Calculating all these individual probabilities would be time-consuming. However, we can use a normal approximation to avoid this. For a large sample size, the binomial distribution can be approximated by a normal distribution.
Using the normal approximation, we can calculate the mean (µ) and standard deviation (σ) for the distribution:
µ = np
σ = sqrt(np(1-p))
Here, n is the total number of births (1,000), and p is the probability of a female birth (0.5).
Then, we can convert the lower bound of 600 girls to a z-score. A z-score tells us how many standard deviations an observation is from the mean.
z = (x - µ) / σ
In this case, x = 600, µ = np = 1000 * 0.5 = 500, and σ = sqrt(np(1-p)) = sqrt(1000 * 0.5 * (1-0.5)) = 15.81.
Calculating the z-score:
z = (600 - 500) / 15.81 = 6.33
Using the z-score table or a statistical calculator, we can find that the probability is extremely close to 1 (very close to 100%) for a z-score of 6.33.
Therefore, in a large hospital, there would be more days when at least 60% of the babies born are girls.
b) Small Hospital:
In a small hospital, the total number of births per year is likely to be lower than in a large hospital. Let's assume there are 100 births per year. We can follow the same steps as with the large hospital to calculate the probability of at least 60% of babies being girls.
Using the formula for the mean and standard deviation:
µ = np = 100 * 0.5 = 50
σ = sqrt(np(1-p)) = sqrt(100 * 0.5 * (1-0.5)) = 5
Calculating the z-score:
z = (60 - 50) / 5 = 2
Using the z-score table or a statistical calculator, we can find that the probability is approximately 0.977 for a z-score of 2. Therefore, the probability of having at least 60% of the babies born being girls in a small hospital is around 0.977.
Comparing the probabilities, we can conclude that in a large hospital, the probability of having at least 60% of babies being girls is significantly higher. Therefore, there would be more days in a large hospital when at least 60% of the babies born are girls compared to a small hospital.
c) It makes a difference:
Based on the calculations above, we can see that the size of the hospital does make a difference. In a large hospital, the probability of having at least 60% of babies being girls is very close to 100%, whereas in a small hospital, this probability is around 97.7%. Thus, the probability is higher in a large hospital compared to a small hospital.