Problem 32
Question
Monthly sales are independent normal random variables with mean 100 and standard deviation 5. (a) Find the probability that exactly 3 of the next 6 months have sales greater than \(100 .\) (b) Find the probability that the total of the sales in the next 4 months is greater than \(420 .\)
Step-by-Step Solution
Verified Answer
(a) The probability that exactly 3 of the next 6 months have sales greater than $100 is \(0.3125\).
(b) The probability that the total sales in the next 4 months are greater than $420 is approximately \(0.0228\).
1Step 1: (Step 1: Calculate probability of sales greater than \(100)
(We first need to find the probability that the sales in one month are greater than \)100. To do this, we need to find the z-score and then use the standard normal table to find the corresponding probability.
The z-score is given by the formula:
\[z = \frac{x - \mu}{\sigma}\]
In this case, \(x = 100\), \(\mu = 100\) and \(\sigma = 5\). So, we have:
\[z = \frac{100 - 100}{5} = 0\]
Now, we'll use the standard normal table to find the probability P(Z > 0). In this case, P(Z > 0) = 0.5. Thus, the probability of sales in one month being greater than $100 is 0.5.
)
2Step 2: (Step 2: Calculate the probability of exactly 3 sales greater than \(100)
(We know that monthly sales are independent, and we have to find the probability that exactly 3 of the next 6 months have sales greater than \)100. This is a binomial problem, and we can use the binomial probability formula:
\[P(X = k) = \binom{n}{k}p^k(1-p)^{n-k}\]
Here, \(n = 6\), \(k = 3\), and \(p = 0.5\). So, we have:
\[P(X = 3) = \binom{6}{3}(0.5)^3(1-0.5)^{6-3} = 20 \times (0.5)^3(0.5)^3 = 0.3125\]
Thus, the probability that exactly 3 of the next 6 months have sales greater than $100 is 0.3125.
)
3Step 3: (Step 3: Summing sales for the next 4 months)
(We need to find the probability that the total sales in the next 4 months is greater than $420. We first find the mean and standard deviation of the total sales for the next 4 months.
As the sales in each month are independent, the mean and standard deviation of the total sales for the next 4 months are:
Mean: \(\mu_{total} = 4 \times \mu = 4 \times 100 = 400\)
Standard deviation: \(\sigma_{total} = \sqrt{4} \times \sigma = 2 \times 5 = 10\)
)
4Step 4: (Step 4: Calculate the probability for total sales greater than \(420)
(Now we have to find the probability that the total sales in the next 4 months is greater than \)420. We first find the z-score for the total sales with \(x = 420\), \(\mu_{total} = 400\), and \(\sigma_{total} = 10\):
\[z = \frac{420 - 400}{10} = 2\]
Now, we'll use the standard normal table to find the probability P(Z > 2). In this case, P(Z > 2) = 0.0228 (approx). Thus, the probability that the total sales in the next 4 months are greater than $420 is approximately 0.0228.
Therefore, the answers are:
(a) The probability that exactly 3 of the next 6 months have sales greater than $100 is 0.3125.
(b) The probability that the total sales in the next 4 months are greater than $420 is approximately 0.0228.
Key Concepts
Normal Random VariablesStandard DeviationBinomial ProbabilityZ-Score
Normal Random Variables
In probability theory, normal random variables are a crucial concept. They're based on the normal distribution, also known as the Gaussian distribution, which is a continuous probability distribution frequently used in statistics and natural sciences. It's recognized by its distinctive bell-shaped curve, symmetric about the mean.
For normal random variables, any two quantities with the same mean and standard deviation will have the same distribution. This property is fundamental because it allows us to model real-world phenomena that tend to cluster around a central value. In the exercise, monthly sales are treated as normal random variables with a mean of 100 and a standard deviation of 5. This indicates that on average, the sales amount to 100 units per month, with some variability captured by the standard deviation.
For normal random variables, any two quantities with the same mean and standard deviation will have the same distribution. This property is fundamental because it allows us to model real-world phenomena that tend to cluster around a central value. In the exercise, monthly sales are treated as normal random variables with a mean of 100 and a standard deviation of 5. This indicates that on average, the sales amount to 100 units per month, with some variability captured by the standard deviation.
Standard Deviation
Standard deviation is a measure that tells us how spread out numbers are in a data set. In simpler terms, it indicates the average distance between the data points and the mean. A small standard deviation means that the data points tend to be close to the mean, while a large one signifies a wide spread of values.
Knowing the standard deviation helps in understanding the variability or volatility of a set of numbers. For example, the textbook exercise highlights that the standard deviation of monthly sales is 5, suggesting that the sales figures typically deviate from the average by 5 units. This measure is crucial when calculating probabilities for normal random variables as it helps in standardizing the data and comparing it against a normalized scale.
Knowing the standard deviation helps in understanding the variability or volatility of a set of numbers. For example, the textbook exercise highlights that the standard deviation of monthly sales is 5, suggesting that the sales figures typically deviate from the average by 5 units. This measure is crucial when calculating probabilities for normal random variables as it helps in standardizing the data and comparing it against a normalized scale.
Binomial Probability
The binomial probability is related to random events which can have one of two outcomes, commonly referred to as 'success' and 'failure'. For instance, flipping a coin has two outcomes: heads or tails. In our context, a 'success' would be a month where sales exceed 100 units.
The binomial probability formula calculates the likelihood of observing a specific number of successes in a fixed number of trials, when each trial is independent and has the same probability of success. In the given exercise, we were asked to find the probability that exactly 3 of the next 6 months have sales greater than 100. This constitutes a binomial probability scenario, where the number of trials is 6 (months), the number of successes we're interested in is 3 (months with sales over 100), and the probability of success in each trial is 0.5.
The binomial probability formula calculates the likelihood of observing a specific number of successes in a fixed number of trials, when each trial is independent and has the same probability of success. In the given exercise, we were asked to find the probability that exactly 3 of the next 6 months have sales greater than 100. This constitutes a binomial probability scenario, where the number of trials is 6 (months), the number of successes we're interested in is 3 (months with sales over 100), and the probability of success in each trial is 0.5.
Z-Score
A z-score, also known as a standard score, quantifies the number of standard deviations a given data point is from the mean. It's a way to measure relative position within a data set and is a key component in standardizing data to make them comparable. When data points follow a normal distribution, z-scores allow us to determine their probabilities using standard normal tables or the normal cumulative distribution function.
In the exercise, to find the probability of monthly sales being greater than 420, a z-score needs to be calculated. It's a crucial step because the z-score transforms the sales value of 420 into a score that can be used to find probabilities on the standard normal distribution. This score then tells us how likely it is to observe a total sales value above 420 in the next 4 months.
In the exercise, to find the probability of monthly sales being greater than 420, a z-score needs to be calculated. It's a crucial step because the z-score transforms the sales value of 420 into a score that can be used to find probabilities on the standard normal distribution. This score then tells us how likely it is to observe a total sales value above 420 in the next 4 months.
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