Questions: At a local company, the length of service for employees is a normal random variable, where the mean length of service is 9.9 years and the standard deviation is 4.1 years. Find the length of service (to the nearest one-tenth of a year) for an employee whose time in service has a 2 -score of 1.46 . A. 15.9 years B. 20.4 years C. 8.5 years D. 18.6 years E. 3.9 years

At a local company, the length of service for employees is a normal random variable, where the mean length of service is 9.9 years and the standard deviation is 4.1 years. Find the length of service (to the nearest one-tenth of a year) for an employee whose time in service has a 2 -score of 1.46 . A. 15.9 years B. 20.4 years C. 8.5 years D. 18.6 years E. 3.9 years
Transcript text: At a local company, the length of service for employees is a normal random variable, where the mean length of service is 9.9 years and the standard deviation is 4.1 years. Find the length of service (to the nearest one-tenth of a year) for an employee whose time in service has a 2 -score of 1.46 . A. 15.9 years B. 20.4 years C. 8.5 years D. 18.6 years E. 3.9 years
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Solution

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Solution Steps

Step 1: Given Information

We are provided with the following parameters for the length of service of employees, which follows a normal distribution:

  • Mean (\( \mu \)) = 9.9 years
  • Standard deviation (\( \sigma \)) = 4.1 years
  • Z-score = 1.46
Step 2: Calculate Length of Service

To find the length of service corresponding to the given z-score, we use the formula:

\[ \text{Value} = \mu + (Z \times \sigma) \]

Substituting the known values:

\[ \text{Value} = 9.9 + (1.46 \times 4.1) \]

Calculating the product:

\[ 1.46 \times 4.1 = 5.986 \]

Now, adding this to the mean:

\[ \text{Value} = 9.9 + 5.986 = 15.886 \]

Step 3: Round to Nearest Tenth

We round the calculated length of service to the nearest one-tenth of a year:

\[ \text{Length of Service} \approx 15.9 \text{ years} \]

Final Answer

The length of service for an employee whose time in service has a z-score of 1.46 is

\[ \boxed{15.9 \text{ years}} \]

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