The average family spends $78 at the grocery store each week with a standard deviation of $25
How much money per week would a family need to spend to be in the 90th percentile? Explain, in context, what this means.
90th percentile = Z = 1.28
1.28 = Z = (score-mean/SD
Plug in the other values to find the score.
To find out how much money per week a family would need to spend to be in the 90th percentile, we can use the concept of z-scores.
A z-score measures how many standard deviations a value is away from the mean. In this case, we need to find the z-score corresponding to the 90th percentile.
The formula to calculate the z-score is:
z = (x - μ) / σ
Where:
- x is the value we want to convert to a z-score (unknown)
- μ is the mean (average) of the data (in this case, $78)
- σ is the standard deviation of the data (in this case, $25)
To find the z-score corresponding to the 90th percentile, we need to find the z-score that represents the point at which 90% of the data is below (to the left of) it. This value corresponds to a cumulative probability of 0.9.
Using a standard normal distribution table or calculator, we can find that the z-score corresponding to a cumulative probability of 0.9 is approximately 1.28.
Now we can substitute the known values into the formula and solve for x:
1.28 = (x - 78) / 25
Rearranging the equation to solve for x:
1.28 * 25 = x - 78
32 = x - 78
x = 32 + 78
x = 110
Therefore, a family would need to spend approximately $110 per week to be in the 90th percentile.
In context, this means that if a family spends $110 or more per week at the grocery store, their spending would be higher than 90% of other families. Only 10% of families would spend more than they do.
To determine the amount of money a family would need to spend per week to be in the 90th percentile, we can use the concept of z-scores.
The 90th percentile is a measure that indicates that 90% of the data falls below a certain value. In this case, we want to find the amount of money per week that only 10% of families spend more than.
To calculate the z-score, we can use the formula:
z = (x - μ) / σ
Where:
- x is the value we want to find the z-score for (the amount of money per week).
- μ is the mean (average) amount of money spent by families per week.
- σ is the standard deviation.
Given that the average family spends $78 per week with a standard deviation of $25, we know that μ = 78 and σ = 25.
To find the z-score corresponding to the 90th percentile, we need to find the z-value such that the area under the normal distribution curve to the left of that z-value is 0.9 (since the 90th percentile means 90% of families spend less than that amount).
Using a standard normal distribution table or a calculator, we find that the z-value for a cumulative area of 0.9 is approximately 1.28.
Now, we can rearrange the z-score formula to solve for x:
x = z * σ + μ
x = 1.28 * 25 + 78
x ≈ 109.2
Therefore, a family would need to spend approximately $109.20 per week to be in the 90th percentile. This means that only 10% of families spend more than this amount on groceries each week.