A police radar gun is used to measure the speeds of cars on a highway. The speeds of cars are normally distributed with a mean of 55 mi/hr and a standard deviation of 5 mi/hr. Roughly what proportion of cars are driving between 60 and 70 mi/hr? Round to the nearest thousandth) Group of answer choices

Answer:

Approximately [tex]0.157[/tex].

Step-by-step explanation:

Let [tex]X[/tex] denote the speed of a car. The question is asking for [tex]P(60 \le X \le 70)[/tex].

That probability is equal to [tex]P(X \le 70)- P(X \le 60)[/tex].

Find each of [tex]P(X \le 60)[/tex] and [tex]P(X \le 70)[/tex] using a [tex]Z[/tex]-table.

Calculate the [tex]Z[/tex] value for [tex]x = 70[/tex] and [tex]x = 60[/tex]:

At [tex]x = 70[/tex], [tex]\displaystyle z = \frac{x - \mu}{\sigma} = \frac{60 - 55}{5} = 3[/tex].

At [tex]x = 60[/tex], [tex]\displaystyle z = \frac{x - \mu}{\sigma} = \frac{70 - 55}{5} = 1[/tex].

In other words: [tex]P(X \le 70) = P(Z \le 3)[/tex] whereas [tex]P(X \le 60) = P(Z \le 1)[/tex].

Look up [tex]P(Z \le 3)[/tex] and [tex]P(Z \le 1)[/tex] on a [tex]Z[/tex]-table. The question is asking that the answer be rounded to the nearest thousandth. Therefore, keep at least more decimal places than that in intermediate values.

[tex]P(Z \le 3) \approx 0.9987[/tex] and [tex]P(Z \le 1) \approx 0.8413[/tex].

Therefore:

[tex]\begin{aligned}P(60 \le X \le 70) &= P(X \le 70)- P(X \le 60) \\ &= P(Z \le 3) - P(Z \le 1) \approx 0.157\end{aligned}[/tex].

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