Question

The proportion of traffic fatalities resulting from drivers with high alcohol blood levels in 1982 was approximately normally distributed among U.S. states, with mean 0.569 and standard deviation 0.068 (U.S. Department of Transportation Traffic Safety Facts 1999). a. What proportion of states would you expect to have more than 65% of their traffic fatali- ties from drunk driving? b. What proportion of deaths due to drunk driving would you expect to be at the 25th percentile of this distribution?

Answer 1

Answer:

a) The proportion of states would you expect to have more than 65% of their traffic fatalities from drunk driving is 0.59518.

b) The 25th percentile of this distribution is a 0.5231 proportion of deaths due to drunk driving.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by

$Z = \frac{X - \mu}{\sigma}$

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the percentile of this measure.

In this problem, we have that:

Mean 0.569 and standard deviation 0.068. So $\mu = 0.569, \sigma = 0.068$

a. What proportion of states would you expect to have more than 65% of their traffic fatali- ties from drunk driving?

This is the value of X when Z has a pvalue of 0.65.

Z has a pvalue of 0.65 between $Z = 0.38$ and $Z = 0.39$. So, we use $Z = 0.385$

$Z = \frac{X - \mu}{\sigma}$

$0.385 = \frac{X - 0.569}{0.068}$

$X - 0.569 = 0.068*0.385$

$X = 0.59518$

The proportion of states would you expect to have more than 65% of their traffic fatalities from drunk driving is 0.59518.

b. What proportion of deaths due to drunk driving would you expect to be at the 25th percentile of this distribution?

This is the value of X when Z has a pvalue of 0.25.

Z has a pvalue of 0.65 between $Z = -0.67$ and $Z = -0.68$. So, we use $Z = -0.675$

$Z = \frac{X - \mu}{\sigma}$

$-0.675 = \frac{X - 0.569}{0.068}$

$X - 0.569 = 0.068*(-0.675)$

$X = 0.5231$

The 25th percentile of this distribution is a 0.5231 proportion of deaths due to drunk driving.

Answer 2

Using the normal distribution principle, the proportion expected to have more than 65% and number expected to be at the 25% percentile are 0.5428 and 0.5232

Given the Parameters :

• Mean, μ = 0.569

• Standard deviation, σ = 0.068

https://brainly.com/question/13684655

• Zscore = (X - μ) ÷ σ

1.) More than 65% of traffic fatalities :

Using a normal distribution table ; Z score for a Pvalue > 0.65 would be : -0.674

Using the Zscore relation :

-0.674 = (X - 0.569) / 0.068

-0.02618 = X - 0.569

X = -0.02618 + 0.569

X = 0.5428

2.) Proportion at the 25 percentile of the distribution :

Using a normal distribution table ; Z score for a Pvalue of 0.25 would be : -0.385

Using the Zscore relation :

-0.385 = (X - 0.569) / 0.068

-0.045832 = X - 0.569

X = -0.02618 + 0.569

X = 0.5232

Therefore, the proportion expected to be at the 25th Percentile is 0.5232.

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