the monthly demand for a certain product is a normal random variable with a mean of 50 and standard deviation 10 a) if you have 60 units on a stock at the beginning of the month, what is the probability you will still have stock remaining at the end of the month? b)what is the probability that the total demand over the next 3 months exceeds 180
to estimate p, the proportion of all newborn babies who are male, the gender of 10,000 newborn babies was noted. if 5106 were male determine a) a 90 percent confidence interval estimate of p b) a 99 percent confidence interval estimate of p?
To answer these questions, we need to use the properties of the normal distribution. The probability distribution of the monthly demand for the product follows a normal distribution with a mean of 50 and a standard deviation of 10.
a) To find the probability of still having stock remaining at the end of the month, we need to find the probability that the monthly demand is less than or equal to 60. This is equivalent to finding the area under the normal curve to the left of 60.
To calculate this probability, we can use a standard normal distribution table or a calculator that provides normal distribution functions. The standard normal distribution (z-distribution) has a mean of 0 and a standard deviation of 1. We can convert the given values to the standard normal distribution by using the formula z = (x - μ) / σ, where x is the value, μ is the mean, and σ is the standard deviation.
In this case, we want to find P(X ≤ 60), where X is the monthly demand. So, we calculate z = (60 - 50) / 10 = 1.
Now, we can use the standard normal distribution table or a calculator to find the probability associated with a z-value of 1. Generally, the table provides the probability up to a certain z-value. For example, if the table gives the probability associated with a z-value of 1 as 0.8413, then P(X ≤ 60) ≈ 0.8413.
b) To find the probability that the total demand over the next 3 months exceeds 180, we need to find the probability that the sum of the monthly demands for three months is greater than 180.
Let Y be the total demand for the next 3 months. The total demand Y is a normal random variable with a mean of 3 * 50 = 150 and a standard deviation of √(3 * 10^2) = 10√3.
We want to find P(Y > 180). Since we know the mean and standard deviation of Y, we can convert this to a standard normal distribution by using the formula z = (Y - μ) / σ, where Y is the value, μ is the mean, and σ is the standard deviation.
For the given problem, z = (180 - 150) / (10√3) ≈ 1.732.
Now, we can use the standard normal distribution table or a calculator to find the probability associated with a z-value of 1.732. This will give us the probability P(Y > 180).
Note that the standard normal distribution table usually provides the probability associated with a z-value up to a certain decimal place. If the table doesn't contain the exact value, you may need to use interpolation or use a calculator that provides normal distribution functions to find the probability.