The mean amount spent by a family of four on food is $500 per month with a standard deviation of $75. Assuming that the food costs are normally distributed, what is the probability that a family spends less than $410 per month? 0.1151 0.8750 0.2158 0.0362
Continuous Probability Distributions
Probability distributions are of two types, which are continuous probability distributions and discrete probability distributions. A continuous probability distribution contains an infinite number of values. For example, if time is infinite: you could count from 0 to a trillion seconds, billion seconds, so on indefinitely. A discrete probability distribution consists of only a countable set of possible values.
Normal Distribution
Suppose we had to design a bathroom weighing scale, how would we decide what should be the range of the weighing machine? Would we take the highest recorded human weight in history and use that as the upper limit for our weighing scale? This may not be a great idea as the sensitivity of the scale would get reduced if the range is too large. At the same time, if we keep the upper limit too low, it may not be usable for a large percentage of the population!
![**Probability of Family Food Expenditure**
In a study of family food expenditures, it is observed that a family of four spends an average (mean) of $500 per month on food, with a standard deviation of $75. Given that the food costs follow a normal distribution, we aim to determine the probability that a family's monthly food expenditure is less than $410.
The probability options provided are:
- 0.1151
- 0.8750
- 0.2158
- 0.0362
**Explanation of the Calculation:**
1. To find the probability, we first need to calculate the Z-score for $410.
\[ Z = \frac{X - \mu}{\sigma} \]
Where:
- \( Z \) is the Z-score.
- \( X \) is the value we are interested in ($410).
- \( \mu \) is the mean ($500).
- \( \sigma \) is the standard deviation ($75).
2. Plugging in the values,
\[ Z = \frac{410 - 500}{75} = \frac{-90}{75} = -1.2 \]
3. We then use the Z-table (standard normal distribution table) to find the probability corresponding to a Z-score of -1.2.
Using the Z-table, the probability (P) corresponding to a Z-score of -1.2 is approximately 0.1151.
Therefore, the probability that a family spends less than $410 per month on food is approximately **0.1151**.
On an educational website, this problem could be used to illustrate how to apply the properties of a normal distribution to real-world situations involving probabilities.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F68e1928e-0b26-44c3-8674-81d02ae0cec4%2Ff7d40193-cac4-4b8a-b4f2-36e4f4b74ca7%2Fn2d94zs_processed.png&w=3840&q=75)
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