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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).

Respuesta :

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) [tex]\bar{x}[/tex]: 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 [tex]\bar{x}[/tex] is a normal distribution with

[tex]\text{Mean} = 2.2\\\text{Standard Deviation} = \displaystyle\frac{\sigma}{\sqrt{n}} = \frac{1.4}{\sqrt{52}} = 0.19[/tex]

b) P(mean is less than 2)

Formula:

[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]

P(x > 610)

[tex]P( \bar{x} < 2) = P( z < \displaystyle\frac{2 - 2.2}{0.19}) = P(z < -1.052)[/tex]

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

[tex]P(\bar{x} < 2) = 0.146[/tex]

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

P(x < 100)

[tex]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)[/tex]

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

[tex]P(x<100) =0.070[/tex]

In the probability, the approximate distribution will be 0.19.

How to calculate the probability

From the given information, the number of accidents per week at a hazardous intersection varies with mean 2.2 and standard deviation 1.4.

Therefore, the approximate distribution of (x-bar) according to the central limit theorem will be:

= 1.4/✓52

= 0.19

The approximate probability that (x-bar) is less than 2 will be:

= P(z < (2-2.2) / 0.19)

= P(z < - 1.052)

Therefore, thus will be 0.146 from the normal distribution table.

Lastly, the probability that there are fewer than 100 accidents at the intersection in a year will be:

= P(x - 100/52)

= P(x < 1.92)

= P(z < 1.92-2.2/0.19)

= P(z < -1.473)

This will be 0.070 from the normal distribution table.

Learn more about probability on:

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