Ask a New Question
Menu

In a recent survey of 655 working Americans ages 25-34, the average weekly amount spent on lunch was 46.43 with standard deviation 2.90. The weekly amounts are approximately bell-shaped.

Estimate the percentage of amounts that are between 40.63 and 52.23.

5 answers

To estimate the percentage of amounts spent on lunch that fall between $40.63 and $52.23, we can use the properties of a bell-shaped (normal) distribution, specifically the empirical rule (68-95-99.7 rule).

Given:

  • Mean (\(\mu\)) = $46.43
  • Standard deviation (\(\sigma\)) = $2.90

First, we can calculate how many standard deviations the values of $40.63 and $52.23 are from the mean:

  1. Calculate the z-score for $40.63:

\[ z_1 = \frac{X - \mu}{\sigma} = \frac{40.63 - 46.43}{2.90} = \frac{-5.80}{2.90} \approx -2.00 \]

  1. Calculate the z-score for $52.23:

\[ z_2 = \frac{X - \mu}{\sigma} = \frac{52.23 - 46.43}{2.90} = \frac{5.80}{2.90} \approx 2.00 \]

Now, we have:

  • \( z_1 \approx -2 \)
  • \( z_2 \approx 2 \)

According to the empirical rule:

  • Approximately 95% of the data lies within 2 standard deviations from the mean in a normal distribution.

Thus, the percentage of amounts spent on lunch that are between $40.63 and $52.23 is approximately 95%.

(b) Estimate the percentage of amounts that are between 37.73 and 55.13

To estimate the percentage of amounts spent on lunch that fall between $37.73 and $55.13, we will again use the properties of a normal distribution.

Given:

  • Mean (\(\mu\)) = $46.43
  • Standard deviation (\(\sigma\)) = $2.90

We can calculate the z-scores for both $37.73 and $55.13.

  1. Calculate the z-score for $37.73:

\[ z_1 = \frac{X - \mu}{\sigma} = \frac{37.73 - 46.43}{2.90} = \frac{-8.70}{2.90} \approx -3.00 \]

  1. Calculate the z-score for $55.13:

\[ z_2 = \frac{X - \mu}{\sigma} = \frac{55.13 - 46.43}{2.90} = \frac{8.70}{2.90} \approx 3.00 \]

Now, we have:

  • \( z_1 \approx -3 \)
  • \( z_2 \approx 3 \)

According to the empirical rule:

  • Approximately 99.7% of the data falls within 3 standard deviations from the mean in a normal distribution.

Therefore, the percentage of amounts spent on lunch that are between $37.73 and $55.13 is approximately 99.7%.

(c) Between what two values will approximately 68% of the amounts be?

To find the two values between which approximately 68% of the amounts will fall in a normal distribution, we can use the empirical rule (68-95-99.7 rule), which states that about 68% of the data lies within 1 standard deviation from the mean.

Given:

  • Mean (\(\mu\)) = $46.43
  • Standard Deviation (\(\sigma\)) = $2.90

To find the range that contains approximately 68% of the amounts, we calculate:

  1. Lower Bound: \[ \text{Lower Bound} = \mu - \sigma = 46.43 - 2.90 \approx 43.53 \]

  2. Upper Bound: \[ \text{Upper Bound} = \mu + \sigma = 46.43 + 2.90 \approx 49.33 \]

Thus, approximately 68% of the amounts spent on lunch will fall between $43.53 and $49.33.