The mean amount spent by a family on food per month is $500 with a standard deviation of $75. Assuming that the food costs are normally distributed, what is the probability that a family spends more than $410 per month? Plesase show formula and calcuation.
Use that table and the Z score formula again.
I hope this helps.
To calculate the probability that a family spends more than $410 per month on food, we can use the standard normal distribution formula and the z-score.
1. Start by calculating the z-score. The z-score measures how many standard deviations an observation is from the mean. It is calculated using the formula:
z = (x - μ) / σ
Where:
- x is the given value ($410),
- μ is the mean amount spent by a family on food per month ($500), and
- σ is the standard deviation ($75).
Plugging in these values, we get:
z = (410 - 500) / 75
= -1.2
2. Look up the z-score in the standard normal distribution table or use a calculator. The table provides the probability associated with that z-score.
The probability associated with a z-score of -1.2 is approximately 0.1151.
3. However, since we want to find the probability of spending MORE than $410 per month, we need to calculate the area under the curve to the right of the z-score. This is equal to 1 minus the probability obtained in step 2.
P(X > 410) = 1 - 0.1151
= 0.8849
Therefore, there is approximately an 88.49% chance that a family spends more than $410 per month on food.
Note: The z-score and the standard normal distribution table assume that the data in question follows a normal distribution.