How can I find the Z score of 0.05?

This is the question:

To estimate the average speed of cars on a specific highway, an investigator collected speed data from a random sample of 75 cars driving on the highway. The sample mean and sample standard deviation are 58 miles per hour and 15 miles per hour, respectively. Construct a 90% confidence interval for the mean speed.

I have the answer for it, and this is the answer:

But, I don't understand why Z0.05 =1.645. On the Standard Normal Distribution Table, P(Z < -1.645) = 0.05. Therefore, Z0.05 should be -1.645 instead of 1.645

• The area under the curve must always be positive. The negative is the $z$ value. What this is saying is that the area under the curve and to the left of the $z$ value is 1.645. – John Douma Apr 21 '18 at 19:39
• So, Z0.05 means the total area under the curve and to the left of 0.05? – Rongeegee Apr 21 '18 at 20:37
• Yes, and due to the symmetry of the curve, it is the same as the area to the right of $Z0.95$. You trim 5% of the area from each side to get 90% of the area under the curve. – John Douma Apr 21 '18 at 21:13
• There is an annoying possibility of confusion in notations such as $Z_{0.05},$ which usually means a z-value that cuts 5% of the probability from the upper tail of the dist'n; that's +1.645. By contrast 'quantile 0.05' cuts 5% prob from the lower tail; that's -1.645. // Software tends to use the 'quantile' notation; printed tables tend to use the 'upper-tail probability' notation. – BruceET Apr 21 '18 at 22:31
• the standard normal distribution table gives you the area under the curve to the left, why does the table give me 0.5199 when z=0.05. – Rongeegee Apr 21 '18 at 22:34

1 Answer

Since the normal distribution is symmetric, the sign of $Z_{\alpha /2}$ is not as important as the fact that 5% of the area under the curve is in each tail of the bell curve. Since your confidence interval is constructed by using $$\bar{x} \pm Z_{\alpha /2} \frac{s}{\sqrt{n}}$$ you will be using both the positive a negative of this Z value.