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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

User Selin
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1 Answer

4 votes

Answer:

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

Explanation:

Given -

Mean
(\\u ) = 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} - \\u }{(\sigma )/(√(n))}< (2 - 2.2 )/((1.4)/(√(52)))) Putting
(Z =\frac{\overline{X} - \\u }{(\sigma )/(√(n))})

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

= 0.1515

User Siddhesh
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