Question

the average time students need to finish a particular test is 70 minutes with a standard deviation of 12 minutes. (assume that these times are normally distributed.) if we want 90% of the students to have sufficient time to finish the test, how much time should we give them? 55 minutes 86 minutes 187 minutes 255 minutes

Explanation:

Step1: Find the z - score

For a 90% confidence level (one - tailed), the z - score \(z\) from the standard normal distribution table is \(z = 1.28\) (since \(P(Z<z)=0.9\)).

Step2: Use the z - score formula

The z - score formula is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 70\) (mean), \(\sigma=12\) (standard deviation), and we want to find \(x\).
Rearrange the formula for \(x\): \(x=\mu + z\sigma\).
Substitute the values: \(x = 70+1.28\times12\).
First, calculate \(1.28\times12 = 15.36\).
Then, \(x=70 + 15.36=85.36\approx86\).

Answer:

86 minutes