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?
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%.
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%.
To find the probability that less than half of the staff members will be male, we need to calculate the cumulative probability of having 0 males, 1 male, 2 males, and so on up to 24 males in a sample of 50.
Let's use the binomial probability formula to calculate the probability for each number of males in the sample:
P(X=k) = (nCk) * p^k * (1-p)^(n-k)
Where:
- P(X=k) is the probability of having k males in the sample
- n is the total sample size (50 in this case)
- k is the number of males in the sample
- p is the probability of selecting a male (0.60 in this case)
Now, let's calculate the cumulative probability by summing up the individual probabilities for each possible number of males less than half (25 males or less):
P(X<25) = P(X=0) + P(X=1) + P(X=2) + ... + P(X=24)
= Σ(P(X=k)) for k=0 to 24
Now, let's calculate each probability and sum them up to find the final result.
To calculate the probability that less than half of the staff members in the random sample will be male, we need to use the binomial distribution.
The binomial distribution is a probability distribution that represents the number of successes in a fixed number of independent Bernoulli trials, where each trial has the same probability of success.
In this case, the probability of success is the probability of selecting a male staff member, which is 60% or 0.60. The probability of failure (selecting a female staff member) is 40% or 0.40.
Let's calculate the probability step by step:
Step 1: Determine the number of trials: In this case, the number of trials is 50 since you have a random sample of 50 staff members.
Step 2: Determine the number of successes: We want to find the probability of having less than half of the staff members as male. So, we need to calculate the probability of having 0, 1, 2, 3, ..., or 24 males (since less than 25 males would be less than half).
Step 3: Calculate the probability of each number of successes using the binomial probability formula:
P(X = k) = (nCr) * p^k * q^(n-k)
Where:
- P(X = k) is the probability of having k successes in n trials
- (nCr) represents the binomial coefficient, which is the number of ways to choose k successes from n trials
- p is the probability of success
- q is the probability of failure (1 - p)
- k is the number of successes
- n is the number of trials
Step 4: Calculate the probability of having less than half males by summing up the probabilities of each number of successes from 0 to 24 (less than 25 males).
Using this approach, you can calculate the final probability.