Finding the confidence interval of a normally distributed sample

Question

Traffic police monitor the speed of vehicles as they travel over a new bridge. The average speed for a sample of 27 vehicles was 91.29 km/h, with the sample standard deviation being 4.94 km/h. We will assume that the speeds are Normally distributed, and the police are interested in the mean speed.

Question:Create a 95 % confidence interval for the mean speed of vehicles crossing the bridge. Give the upper and lower bounds to your interval, each to 2 decimal places.

Attempt: The photo I uploaded. I don't know why is it wrong? I can't seem to find the SE(x) so I use the SD(x). Probably that is the mistake? Well I can't find the SE(x), anybody knows how?

Answer 1

I don't see anything wrong with your method. Here is a computation (in R) with essentially no round-off error.

 a = 91.29; s = 4.94; n = 27; pm = c(-1,1); t.c = qt(.975,n-1)
 a + pm*t.c*s/sqrt(n)
 ## 89.3358 93.2442
 t.c
 ## 2.055529

$SE(\bar X) = s/\sqrt{n} = 0.9507034.$