The mean number of accidents a week at a company is 6.4 with a standard deviation of 1.5. What proportion of weeks would you expect to have less than 5 accidents?

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:

Given that μ = 6.4, σ = 1.5:

For x < 5 accidents:

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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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

i = 2 / - 5

step-by-step explanation:

evaluation of ( 2 - 4 ) ( p + q ) when p = 2 and q = 5 i, by putting values of p and q,