BIOD 121 /BIOD 121 Statistics Test 5 /Questions Answers/Latest Updated Solution

BIOD 121 Statistics Test 5: Questions Answers

1. Describe, in your own words the sampling distribution of x̄.
This is the sampling distribution of all possible values of the sample mean, x̄, computed from a sample of size n.
It is important that the same sample size, n, be used for all of the samples.
2. For the sampling distribution of x̄, is it necessary that the samples used in the distribution be all of the same
size?
It is important that all of the samples be of the same size, n.
3. What is E(x̄)? What is E(x̄) equal to?
E(x̄) is the expected value of x̄. E(x̄) is equal to the population mean, µ.
4. Suppose you take a random sample and find the sample mean, x̄. Based on the previous problem, can you
say that x̄ is equal to the mean of the entire population µ?
The random sample mean, x̄ is not, in general, equal to the population mean, µ. x̄ only estimatesµ. The previous
problem tells us that the expected value of x̄ is equal to µ. It does not say that the sample mean of one sample is
equal to µ.
5. A nursing home has 2000 patients. You sample 170 patients. Can you use the infinite standard deviation
formula?
In order to use infinite standard deviation formula, we must have:



We have:



6. Suppose that you take a sample of size 40 from a population that is normally distributed. Can the sampling
distribution of x̄ be approximated by a normal probability distribution? What if the population was not
normally distributed, can the sampling distribution of x̄ be approximated by a normal probability
distribution?
Since the sample size is greater than 30, the sampling distribution of x̄ can be approximated by a normal
probability distribution in both cases (both for normally distributed populations and for populations that are not
normally distributed).
7. Suppose that in a certain large city, the mean 100 meter dash time for male high school seniors was reported
to be 13.3 seconds with a standard deviation of 1.9 seconds. We may assume that the population is normally
distributed.
a) What is the probability that 20 male high school seniors selected at random will have a mean 100 meter dash
time of 12.5 seconds or less?
a) We calculate the standard deviation of the sample distribution:

, Calculate the z-score:




So, we want to find P(Z < -1.88) on the standard normal probability distribution table. We find that, P(Z < -
1.88) = .03005.
Therefore, there is a 0.3005 probability that a simple random sample of 20 male high school seniors will run a
100 meter dash with a mean faster than 12.5 seconds. (In other words, there is a 0.3005 probability that these
seniors will have a mean time of 12.5 seconds or less.)
b) What is the probability that 40 male high school seniors selected at random will have a mean 100 meter dash
time of 12.5 seconds or less?
We calculate the standard deviation of the sample distribution:




Calculate the z-score:




So, we want to find P(Z < -2.66) on the standard normal probability distribution table. We find that, P(Z < -
2.66) = .00391.
Therefore, there is a 0.00391 probability that a simple random sample of 40 male high school seniors will run a
100 meter dash with a mean faster than 12.5 seconds.
c) What is the probability that 35 male high school seniors selected at random will have a mean 100 meter dash
time between 13 seconds and 14 seconds?
We calculate the standard deviation of the sample distribution:




Calculate the z-score for 13 seconds:




Calculate the z-score for 14 seconds:




So, we want to find P(-.93< Z < 2.18) on the standard normal probability distribution table. Recall that P(-.93< Z <
2.18)=P(Z<2.18)-P(Z<-.93)=.98537-.17619=.80918.