To solve this problem, we can use the binomial probability formula.
The probability of success (male) is 0.6, and the probability of failure (female) is 0.4.
We are interested in finding the probability that less than half (25 or fewer) out of the 50 staff members will be male.
We can use the binomial probability formula to calculate this probability:
P(X ≤ 25) = Σ [from x = 0 to 25] (50Ck) * (0.6)^k * (0.4)^(50-k)
Using a binomial probability calculator, or a calculator with a binomial probability function, we find that:
P(X ≤ 25) ≈ 0.3283
Therefore, the probability that less than half of the staff members will be male in a random sample of 50 is approximately 0.3283, or 32.83%.
In a particular hospital, 60% of the staff is male and 40% female. In a random sample of 50 staff members, what is the probability that less than half will be male?
3 answers
f the mean and standard deviation of serum iron values for healthy men are 120 and 15 micrograms per 100 ml, what is the probability that a random sample of 50 healthy men will yield a mean more than 125 micrograms per 100 ml?
To solve this problem, we can use the Central Limit Theorem. According to the Central Limit Theorem, the distribution of sample means of a sufficiently large sample size (n > 30) will be approximately normally distributed, regardless of the shape of the population distribution, as long as the population standard deviation is known.
Given that the mean (μ) of the serum iron values for healthy men is 120 micrograms per 100 ml and the standard deviation (σ) is 15 micrograms per 100 ml, we can set up the standard error of the mean (SE) as:
SE = σ / √n
where n is the sample size. In this case, n = 50.
SE = 15 / √50 ≈ 2.121
To find the probability that a random sample of 50 healthy men will yield a mean more than 125 micrograms per 100 ml, we need to calculate the z-score associated with this value and find the corresponding probability using the standard normal distribution table.
z = (x - μ) / SE
where x is the sample mean.
z = (125 - 120) / 2.121 ≈ 2.358
Using the z-score of 2.358 and the standard normal distribution table, we find that the probability of obtaining a z-score greater than 2.358 is approximately 0.009 or 0.9%.
Therefore, the probability that a random sample of 50 healthy men will yield a mean more than 125 micrograms per 100 ml is approximately 0.009 or 0.9%.
Given that the mean (μ) of the serum iron values for healthy men is 120 micrograms per 100 ml and the standard deviation (σ) is 15 micrograms per 100 ml, we can set up the standard error of the mean (SE) as:
SE = σ / √n
where n is the sample size. In this case, n = 50.
SE = 15 / √50 ≈ 2.121
To find the probability that a random sample of 50 healthy men will yield a mean more than 125 micrograms per 100 ml, we need to calculate the z-score associated with this value and find the corresponding probability using the standard normal distribution table.
z = (x - μ) / SE
where x is the sample mean.
z = (125 - 120) / 2.121 ≈ 2.358
Using the z-score of 2.358 and the standard normal distribution table, we find that the probability of obtaining a z-score greater than 2.358 is approximately 0.009 or 0.9%.
Therefore, the probability that a random sample of 50 healthy men will yield a mean more than 125 micrograms per 100 ml is approximately 0.009 or 0.9%.