To find the probability that the sample mean is greater than 18, we need to calculate the z-score and then use the z-table.
The z-score is given by:
z = (x - u) / (σ / sqrt(n))
where x is the value we are comparing to (18 in this case), u is the population mean (17), σ is the population standard deviation (-6), and n is the sample size (36).
Plugging in the values, we get:
z = (18 - 17) / (-6 / sqrt(36))
= 6 / (-6 / 6)
= 6 / -1
= -6
Next, we look up the corresponding probability in the z-table for -6. From the z-table, we find that the probability associated with a z-score of -6 is very close to 0. Therefore, the probability that the sample mean is greater than 18 is approximately 0.
None of the given options match this result, so it seems that there may be an error in the provided information or answer choices.
A random sample of size 36 is taken from a population with mean u 17 and standard deviation -6. The probability that the sample mean is greater than 18 is
05675
0.8413
0.4325
0.1587
1 answer