Answer:
0.17619
Step-by-step explanation:
We solve using z score formula
= z = (x-μ)/σ,
where x is the raw score = 5 accidents
μ is the population mean = 6.4 accidents
σ is the population standard deviation = 1.5
z = 5 - 6.4/1.5
= -1.4/1.5
= -0.93333
Approximately to 2 decimal places = -0.93
Determining the probability value from Z-Table:
P(x< 5) = P(z = -0.93) = 0.17619
Converting this to percentage =
The probability of weeks you would expect to have less than 5 accidents is
0.17619
We are asked in the question:
What proportion of weeks would you expect to have less than 5 accidents?
Proportion is also the same as:
Therefore, the proportion of weeks that you expect to have less than 5 accidents = 0.17619
17.62% of the accidents would be less than 5 accidents.
The z score is used to determine by how many standard deviations the raw score is above or below the mean. The z score is given by:
[tex]z=\frac{x-\mu}{\sigma} \\\\where\ \mu=mean,x=raw\ score,\sigma=standard\ deviation[/tex]
Given that μ = 6.4, σ = 1.5:
For x < 5 accidents:
[tex]z = \frac{5-6.4}{1.5}=-0.93[/tex]
From the normal distribution table, P(x < 5) = P(z < -0.93) = 0.1762 = 17.62%
17.62% of the accidents would be less than 5 accidents.
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