Consider the approximately normal population of heights of male college students with mean ì = 69 inches and standard deviation of ó = 4.6 inches. A random sample of 25 heights is obtained
(e) Find P(x > 70). (Give your answer correct to four decimal places.)
(f) Find P(x < 67). (Give your answer correct to four decimal places.)
Not sure how to work these?
Does x = raw score or mean?
If raw score:
Z = (score-mean)/SD
If mean:
Z = (score-mean)/SEm
SEm = SD/√n
In both cases, find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability of Z.
Find the reciprocal of the following number:
-21/23
-21/23=2.50=.9938 or is it 70-69=1over4.6/sqrt25=.92=.8212 for the first one
the second part 67-69=-2over 4.6/sqrt25=.92, -2/.92=-2.17=.0150
To solve these questions, we need to use the concept of z-scores and the standard normal distribution table.
Z-score is a measure of how many standard deviations an individual data point is away from the mean. We can calculate the z-score using the formula:
Z = (x - μ) / σ
Where:
Z = z-score
x = data point
μ = mean
σ = standard deviation
Now let's solve the questions step by step:
(e) P(x > 70):
1. Calculate the z-score:
Z = (70 - 69) / 4.6 = 0.2174
2. Look up the probability associated with the z-score on the standard normal distribution table. In this case, we are interested in finding the probability of the area to the right of the z-score.
The z-score of 0.2174 corresponds to an area of 0.5871.
3. Subtract the area found from 1 to get the probability of x being greater than 70:
P(x > 70) = 1 - 0.5871 = 0.4129
So, P(x > 70) is approximately 0.4129.
(f) P(x < 67):
1. Calculate the z-score:
Z = (67 - 69) / 4.6 = -0.4348
2. Look up the probability associated with the z-score on the standard normal distribution table. In this case, we are interested in finding the probability of the area to the left of the z-score.
The z-score of -0.4348 corresponds to an area of 0.3325.
So, P(x < 67) is approximately 0.3325.
I hope this explanation helps you understand how to solve these types of problems.