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 a normal distribution. Let "x-bar" be the mean number of accidents per week at the intersection during a year (52 weeks). What is the approximate probability that there are fewer than 100 accidents in a year? (Hint: Restate this event in terms of "x-bar")

Answer 1

Answer:

The approximate probability that there are fewer than 100 accidents in a year = .9251

Step-by-step explanation:

Given -

Mean $(\nu )$ =2.2

Standard deviation $\sigma$ = 1.4

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

Then $\overline{X}$ = $\frac{100}{52}$ = 1.92

the approximate probability that there are fewer than 1.92 accidents per week  in a year

[Z   = $\frac{\overline{X} - \nu }{\frac{\sigma}{\sqrt{n}}}$ ]

=  $P(\overline{X} < 1.92 )$  = ($P(\frac{\overline{X} - \nu }{\frac{\sigma}{\sqrt{n}}}< \frac{1.92 - 2.2 }{\frac{1.4}{\sqrt{52}}})$

 = P( $Z < -1.442$ )   ( Using Z table)

=  .9251