a. Find the probability that an individual distance is
greater than 214.30 cm
We find for the value of z score using the formula:
z = (x – u) / s
z = (214.30 – 205) / 8.3
z = 1.12
Since we are looking for x > 214.30 cm, we use the
right tailed test to find for P at z = 1.12 from the tables:
P = 0.1314
b. Find the probability that the mean for 20 randomly
selected distances is greater than 202.80 cm
We find for the value of z score using the formula:
z = (x – u) / s
z = (202.80 – 205) / 8.3
z = -0.265
Since we are looking for x > 202.80 cm, we use the
right tailed test to find for P at z = -0.265 from the tables:
P = 0.6045
c. Why can the normal distribution be used in part (b),
even though the sample size does not exceed 30?
I believe this is because we are given the population
standard deviation sigma rather than the sample standard deviation. So we can
use the z test.
You will get a number with a decimal: 22.360679775.
Answer:
x= 7/4
<u />
Step-by-step explanation:
Find the LCM for x-2,x+3,x^2+x-6 which is (x-2) (x+3)
then multiply
<span>28 - (8+4)*(4-2)
= 28 - 32 * 2
= 28 - 64
= -36
</span>
Answer:
A D and E
Step-by-step explanation:
