If the probability of a newborn child being female is 0.5, find the probability that in 100 births, 55 or more will be female. Use the normal distribution to approximate the binomial distribution.

54.5-50/50 = 4.5/5 = 0.9 z= 0.1841

I don't understand how mean got 54.5 or the equation 54.5-10/50 = 0.8 z=0.1841. I think mean (n)(p) = (100)(.5) = 50 standard deviation (p)(p)(q) square root = 5.

Ah, probabilities and newborns, what a combination! Let's dive in and see what we can do here.

To tackle this problem, it's time to bring out the big guns - the normal distribution! We can use it to approximate the binomial distribution in this case.

First, let's find the mean (μ) and standard deviation (σ) of our binomial distribution.

The mean (μ) of a binomial distribution is given by:
μ = n * p
where n is the number of trials (100 births) and p is the probability of success (in this case, the probability of a newborn child being female which is 0.5).

So, μ = 100 * 0.5 = 50.

The standard deviation (σ) of a binomial distribution is given by:
σ = sqrt(n * p * (1 - p))

So, σ = sqrt(100 * 0.5 * (1 - 0.5)) = sqrt(100 * 0.5 * 0.5) = sqrt(25) = 5.

Now, we can use the normal distribution to approximate the binomial distribution. We need to find the probability that 55 or more births will be female, which is equivalent to finding the probability that a random variable following the normal distribution with mean 50 and standard deviation 5 will be greater than or equal to 55.

To do this, we can standardize our random variable using the Z-score formula:
Z = (X - μ) / σ
where X is the value we want to find the probability for.

So, Z = (55 - 50) / 5 = 1.

Now, we need to find the probability of Z being greater than or equal to 1. We can consult a standard normal distribution table to find this probability.

Looking it up, we find that the probability is approximately 0.1587.

Therefore, using the normal distribution to approximate the binomial distribution, the probability that in 100 births, 55 or more will be female is approximately 0.1587.

Remember, this is just an approximation, but I hope it helps lighten up the probability path for you!

To find the probability that in 100 births, 55 or more will be female, we can approximate the binomial distribution using the normal distribution.

Step 1: Calculate the mean and standard deviation of the binomial distribution.
The mean (μ) of a binomial distribution is equal to n * p, where n is the number of trials and p is the probability of success.
In this case, the mean (μ) is 100 * 0.5 = 50.

The standard deviation (σ) of a binomial distribution is equal to √(n * p * (1 - p)).
In this case, the standard deviation (σ) is √(100 * 0.5 * (1 - 0.5)) = √(25) = 5.

Step 2: Standardize the values.
To use the normal distribution, we need to standardize the values by subtracting the mean (μ) and dividing by the standard deviation (σ).

For 55 or more female births:
Z = (x - μ) / σ
Z = (55 - 50) / 5
Z = 5 / 5
Z = 1

Step 3: Calculate the probability using the standard normal distribution table.
The standard normal distribution table provides the probabilities corresponding to different Z-scores.

The probability of having 55 or more female births can be calculated by subtracting the probability of having 54 or fewer female births from 1.
P(X ≥ 55) = 1 - P(X ≤ 54)

Referencing the standard normal distribution table, we find the probability corresponding to a Z-score of 1 is approximately 0.8413.
P(X ≥ 55) = 1 - 0.8413 ≈ 0.1587

Therefore, the probability that in 100 births, 55 or more will be female is approximately 0.1587.

Mean = np

Standard deviation = √npq
Note: q = 1 - p

n = 100
p = 0.5
q = 0.5

Once you find the mean and standard deviation, use z-scores:
z = (x - mean)/sd
(x = 55)

Check a z-table to find the probability using the z-score you calculated.