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The average miles per gallon of a particular automobile model are approximately normally distributed with a given mean [tex]\mu = 43.8[/tex] miles per gallon and standard deviation [tex]\sigma = 5.1[/tex] miles per gallon. What percentage of the automobiles have an average miles per gallon between 38.7 miles per gallon and 48.9 miles per gallon?

A. 68%
B. 75%
C. 95%
D. 100%

Sagot :

To determine the percentage of automobiles that have an average miles per gallon (mpg) between 38.7 mpg and 48.9 mpg, we will use the properties of the normal distribution. Here are the step-by-step details:

1. Identify the given parameters:
- Mean ([tex]$\mu$[/tex]): 43.8 mpg
- Standard deviation ([tex]$\sigma$[/tex]): 5.1 mpg
- Lower bound: 38.7 mpg
- Upper bound: 48.9 mpg

2. Calculate the z-scores for the lower and upper bounds:

The z-score formula is:
[tex]\[ z = \frac{X - \mu}{\sigma} \][/tex]

- For the lower bound (38.7 mpg):
[tex]\[ z_{\text{lower}} = \frac{38.7 - 43.8}{5.1} = \frac{-5.1}{5.1} = -1 \][/tex]

- For the upper bound (48.9 mpg):
[tex]\[ z_{\text{upper}} = \frac{48.9 - 43.8}{5.1} = \frac{5.1}{5.1} = 1 \][/tex]

3. Find the cumulative probabilities for these z-scores:

Using the z-table or standard normal cumulative distribution function:

- The cumulative probability for [tex]\(z = -1\)[/tex] is approximately 0.1587.
- The cumulative probability for [tex]\(z = 1\)[/tex] is approximately 0.8413.

4. Calculate the probability that the average mpg is between the lower and upper bounds:

[tex]\[ P(38.7 \leq X \leq 48.9) = P(z_{\text{lower}} \leq Z \leq z_{\text{upper}}) = \Phi(z_{\text{upper}}) - \Phi(z_{\text{lower}}) \][/tex]

[tex]\[ P(-1 \leq Z \leq 1) = 0.8413 - 0.1587 = 0.6826 \][/tex]

5. Convert the probability to a percentage:

[tex]\[ \text{Percentage} = 0.6826 \times 100\% = 68.26\% \][/tex]

Therefore, approximately 68.27% of the automobiles have an average miles per gallon between 38.7 mpg and 48.9 mpg. Thus, the correct answer is:

[tex]\[ \boxed{68\%} \][/tex]