Answer:
The mean is usual
Step-by-step explanation:
We have that the mean (m) is equal to 21, the standard deviation (sd) = 1.31 and the sample size (n) = 64
They ask us for P (x <20.5)
For this, the first thing is to calculate z, which is given by the following equation:
z = (x - m) / (sd / (n ^ 1/2))
We have all these values, replacing we have:
z = (20.5 - 21) / (1.31 / (64 ^ 1/2))
z = -3.05
With the normal distribution table (attached), we have that at that value, the probability is:
P (z <-3.05) = 0.0002
The mean is usual because P (x> 20.5) = 1 - P (x <20.5) = 1 - 0.0002 = 0.9998 is a fairly high probability.
No, because 40 x 10 is not 4000, but is is 400. It is 100x greater not 10x
Answer: no cheating on the state test
Step-by-step explanation:
Answer:
We can claim with 95% confidence that the proportion of executives that prefer trucks is between 19.2% and 32.8%.
Step-by-step explanation:
We have a sample of executives, of size n=160, and the proportion that prefer trucks is 26%.
We have to calculate a 95% confidence interval for the proportion.
The sample proportion is p=0.26.
The standard error of the proportion is:
The critical z-value for a 95% confidence interval is z=1.96.
The margin of error (MOE) can be calculated as:

Then, the lower and upper bounds of the confidence interval are:

The 95% confidence interval for the population proportion is (0.192, 0.328).
We can claim with 95% confidence that the proportion of executives that prefer trucks is between 19.2% and 32.8%.
11, 12, 13, 14, 15, 16, 17, 18, 19 Sorry if it doesn't help, I'm being too obvious