According to a company’s website, the top 10% of the candidates who take the entrance test will be called for an interview. The reported mean and standard deviation of the test scores are 65 and 9, respectively. If test scores are normally distributed, what is the minimum score required for an interview? (You may find it useful to reference the z table. Round your final answer to 2 decimal places.)
To find the minimum score required for an interview, we need to determine the cutoff score that corresponds to the top 10% of test scores.
First, we need to find the z-score corresponding to the top 10%. The z-score can be found using the z-table.
The z-score can be calculated using the formula:
z = (x - μ) / σ
Where:
x = raw score
μ = mean
σ = standard deviation
Since we want the top 10%, we are looking for a z-score that corresponds to a cumulative probability of 0.90.
Using the z-table, we find that a z-score of 1.28 corresponds to a cumulative probability of 0.90.
Now we can solve for the minimum score required for an interview.
1.28 = (x - 65) / 9
Simplifying the equation:
1.28 * 9 = x - 65
11.52 = x - 65
x = 11.52 + 65
x = 76.52
Therefore, the minimum score required for an interview is 76.52 (rounded to 2 decimal places).
To find the minimum score required for an interview, we need to find the score that corresponds to the top 10% of the candidates.
First, we need to find the z-score for the top 10% of the distribution.
Using the z-table, we find that the z-score corresponding to the top 10% is approximately 1.28.
Next, we need to find the raw score corresponding to this z-score.
We can use the following formula:
z = (X - μ) / σ
where:
z = the z-score
X = the raw score
μ = the mean
σ = the standard deviation
Rearranging the formula to solve for X, we have:
X = (z * σ) + μ
Plugging in the given values:
X = (1.28 * 9) + 65
X = 11.52 + 65
X = 76.52
Therefore, the minimum score required for an interview is approximately 76.52 when rounded to 2 decimal places.
To find the minimum score required for an interview, we need to calculate the z-score corresponding to the top 10% cutoff point.
First, we need to find the z-score using the formula:
z = (x - μ) / σ
Where:
z is the z-score
x is the cutoff score we want to find
μ is the mean of the distribution (65)
σ is the standard deviation of the distribution (9)
To find the z-score corresponding to the top 10%, we need to look for the z-value in the z-table that corresponds to a cumulative probability of 0.9. This corresponds to the highest 10% of the distribution.
Looking up this value in the z-table, we find that the z-score corresponding to a cumulative probability of 0.9 is approximately 1.28.
Now we can rearrange the formula to solve for x:
x = z * σ + μ
Plugging in the values, we have:
x = 1.28 * 9 + 65
Calculating this expression, we find:
x ≈ 76.52
Therefore, the minimum score required for an interview is approximately 76.52 when rounded to 2 decimal places.