Question

The number of accidents per week at a hazardous intersection varies with mean 2.2 and standard deviation of 1.4. The distribution takes only integers and most definitely not normal. Let x-bar be the mean of accidents per week at the intersection during one year (52 weeks). What is the approximate probability that x-bar is less than 2

Answer 1

Answer:

Approximate probability that $\overline{X}$ is less than 2 = 0.1515

Step-by-step explanation:

Given -

Mean $(\nu )$ = 2.2

Standard deviation $(\sigma )$ = 1.4

Sample size ( n ) = 52

Let  $\overline{X}$ be the mean of accidents per week at the intersection during one year (52 weeks) .

probability that $\overline{X}$ is less than 2 =

$P(\overline{X}< 2)$  = $P(\frac{\overline{X} - \nu }{\frac{\sigma }{\sqrt{n}}}< \frac{2 - 2.2 }{\frac{1.4}{\sqrt{52}}})$   Putting $(Z =\frac{\overline{X} - \nu }{\frac{\sigma }{\sqrt{n}}})$

                 = $P(Z< - 1.03)$    ( Using Z table )

                 = 0.1515