Stat 226 SI:
3/5/13
Supplemental Instruction
IowaStateUniversity / Leader: / Carly
Course: / Stat 226
Instructor: / (Various)
Date: / 3/5/13
  • Exam 2 Material Review
  • Chapter 13: “Samples and Surveys”
  • Accuracy vs. Precision
  • The meaning of and how it is different from and μ
  • Sampling Strategies
  • SRS
  • Voluntary
  • Convenience
  • Stratified
  • Cluster
  • Bias in sampling
  • Shape, center, spread in histograms
  • Chapter 14: “Sampling Dist. of the Sample Mean and the Central Limit Theorem”
  • Shape, center, spread for the dist. of
  • 3 Cases
  • X~N(μ, σ2)
  • ~N
  • X is symmetrically distributed and bell-shaped (but not normal)
  • ~approx. N
  • X is not normal, nor symmetric (skewed, multimodal, etc.)
  • ~approx. N if n is sufficiently large (n ≥ 30)
  • Use of the CLT
  • Standard Deviation of  “Standard Error”
  • Generic Problems for Sampling Distribution of

1.) Variable X is normally distributed with a mean of 50 and standard deviation of 6.

a.) What can we say about the shape of the sampling distribution of ?

is normally distributed because X is normally distributed.

b.) Do we need to use the CLT to comment on the shape of the dist. of ?

No, we do not need to use the CLT because X follows a normal dist.

c.) What are the mean and standard error of the sampling distribution of ?

SE() = (A value for “n” was not given.)

2.) Variable X is not normally distributed, but it is symmetric and bell-shaped. It has a mean of 50 and a standard deviation of 6.

a.)What can we say about the sampling distribution of ?

is approximately normally distributed.

b.)Does the standard error of change if we go from a normal sampling

distribution of to one that is only approximately normal? If so, how?

No, the standard error is still equal to .

c.)Is the mean for a distribution of X different from the mean of a sampling distribution of ? If so, when?

Regardless of the shape of the distribution of X, the sample mean will follow a normal distribution with a mean , according to the CLT.

  • A farmer in Iowa owns an apple orchard. He claims that the number of apples per tree on his apple orchard is normally distributed with a mean of 100 apples and a standard deviation of 20 apples. Assume a sample size of 300.

What is the shape of the distribution of X?

X is normally distributed.

What is the shape of the sampling distribution of ?

is normally distributed because X is normally distributed.

X /
Mean / Standard Deviation / Mean / Standard Error
μ = 100 / σ = 20 / 100 / 1.155

Can we make conclusions about the probability distribution of X?

Yes, because X is normally distributed.

What is the probability of “obtaining” a single tree with more than 115 apples?

P(X > 115)

= 1 – P(X < 115)

= 1– P

= 1 – P(Z < 0.75)

= 1 – 0.7734

= 0.2266

= 22.66%

What is the probability of obtaining a sample mean greater than or equal to 103 apples?

P(≥ 103)

= 1 – P( ≤ 103)

= 1 – P

= 1 – P(Z ≤ 2.60)

= 1 – 0.9953

= 0.0047

= 0.47%