Question

The number of accidents per week at a hazardous intersection varies with mean 2.2 and standard deviation 1.4. This distribution takes only whole-number values, so it is certainly not Normal.
(a) Let (x-bar) be the mean number of accidents per week at the intersection during a year (52 weeks). What is the approximate distribution of (x-bar) according to the central limit theorem?
(b) What is the approximate probability that (x-bar) is less than 2?
(c) What is the approximate probability that there are fewer than 100 accidents at the intersection in a year? (Hint: Restate this event in terms of (x-bar).

Answer 1

Answer:

a) Normal Distribution

b) 0.146

c) 0.070

Step-by-step explanation:

We are given the following information in the question:

The number of accidents per week at a hazardous intersection varies with mean 2.2 and standard deviation 1.4

a) $\bar{x}$: mean number of accidents per week at the intersection during a year (52 weeks)

According to central limit theorem, as the sample size becomes larger, the distribution of mean approaches a normal distribution.

Since we have a large sample, the approximate distribution of $\bar{x}$ is a normal distribution with

$\text{Mean} = 2.2\\\text{Standard Deviation} = \displaystyle\frac{\sigma}{\sqrt{n}} = \frac{1.4}{\sqrt{52}} = 0.19$

b) P(mean is less than 2)

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

P(x > 610)

$P( \bar{x} < 2) = P( z < \displaystyle\frac{2 - 2.2}{0.19}) = P(z < -1.052)$

Calculation the value from standard normal z table, we have,  

$P(\bar{x} < 2) = 0.146$

c) P(fewer than 100 accidents at the intersection in a year)

P(x < 100)

$P( \bar{x} < \frac{100}{52}) =P(\bar{x} < 1.92) =P(z < \displaystyle\frac{1.92 - 2.2}{0.19}) = P(z < -1.473)$

Calculation the value from standard normal z table, we have,  

$P(x$