Final answer:
Using the normal distribution properties, we estimate that 16.0% of amounts were less than $41.52, 2.5% were greater than $49.89, and 34.1% were between $44.31 and $47.10.
Step-by-step explanation:
To estimate the percentage of working Americans ages 25-34 who spent different amounts on lunch, we will use the given average and standard deviation. Since the distribution is said to be approximately bell-shaped, we can use the properties of the normal distribution for our calculations. In this case, the average, also known as the mean, is $44.31, and the standard deviation is $2.79.
Part (a)
To find the percentage of amounts less than $41.52, we first calculate the z-score as follows:
Z = (X - μ) / σ = ($41.52 - $44.31) / $2.79 = -0.9964
Looking at a z-score table, a z-score of -0.9964 corresponds to approximately 16.0%. So, the percentage of amounts less than $41.52 is 16.0%.
Part (b)
For amounts greater than $49.89, we calculate the z-score in a similar manner:
Z = (X - μ) / σ = ($49.89 - $44.31) / $2.79 = 1.9964
This corresponds to a lower tail percentage of approximately 97.5%, meaning that the percentage greater than $49.89 is 100% - 97.5% = 2.5%.
Part (c)
To find the percentage of amounts between $44.31 and $47.10:
We calculate the z-score for $47.10 as follows:
Z = (X - μ) / σ = ($47.10 - $44.31) / $2.79 = 1.0043
A z-score of 1.0043 corresponds to about 84.1%. The percentage between the mean and z-score of 1.0043 is 84.1% - 50% (since the mean is the 50th percentile) = 34.1%.