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Answer:
The correct option is;
A. 0.17
Step-by-step explanation:
The given information are;
The mean duration it takes a cashier to process an order, μ = 276 seconds
The standard deviation from the mean, σ = 38 seconds
The z-score of the order processing time of x = 240 seconds is given as follows;
[tex]Z=\dfrac{x-\mu }{\sigma }[/tex]
Therefore;
[tex]Z=\dfrac{240-276 }{38 } \approx -0.9474[/tex]
The probability
P(z = -0.9474) = 0.17361
Therefore, the proportion of orders which are processed in less than 240 is approximately 0.17361 or 0.17 to two places of decimal.
The closest to the proportion of orders that are processed in less than 240 seconds is 17%
The given parameters;
mean of the number of the distribution, M = 276 seconds
standard deviation of the distribution, d = 38 seconds
In a normal distribution curve, 1 standard deviation below the mean is given as follows;
1 standard deviation below mean = 16% = M - d
= 276 s - 38 s = 238 s
the mean of distribution = 50% = 276 s
Let the proportion less than 240 s = x%
238 s --------------------------- 16%
240 s ---------------------------- x%
276 s ------------------------------- 50%
Use interpolation method, to determine the value of x;
[tex]\frac{240 - 238}{276 - 238} = \frac{x- 16}{50-16} \\\\\frac{2}{38} = \frac{x- 16}{34} \\\\\frac{1}{19} = \frac{x- 16}{34} \\\\19(x-16) = 34\\\\19x - 304 = 34\\\\19x = 338\\\\x = \frac{338}{19} \\\\x = 17.78\ \%[/tex]
Thus, the closest to the proportion of orders that are processed in less than 240 seconds is 17%
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