What is the lowest value of the sample mean in this sampling distribution

Learning Outcomes

• Estimate the probability of an event using a normal model of the sampling distribution.

Example

Surprising Heights for Individual Basketball Players

Suppose we have a population of adult male basketball players and we know their heights: the mean height is μ = 190 cm and the standard deviation of their heights is σ = 7.2 cm. The heights are normally distributed, which is often the case with body measurements.

Would it be surprising to find a randomly chosen player from this population with a height of 195 cm?

We can answer this question by computing the probability that a randomly chosen player from this population has height greater than 195 cm. To carry out the analysis, lets use X to denote the height of a randomly chosen individual from this population. Since heights are normally distributed, we can convert heights to z-scores and use our simulation to find the probability P(X > 195).

1. Convert the interval X > 195 to an interval of z-scores.Recall that the z-score of an X-value is the number of standard deviations that value is away from the mean. The formula is

   [latex]Z\text{}=\text{}\frac{X-μ}{σ}[/latex]

   So the z-score of X = 195 is

   [latex]\frac{X-μ}{σ}\text{}=\text{}\frac{195-190}{7.2}\text{}=\text{}\frac{5}{7.2}\text{}=\text{}0.69[/latex]

   That means that the interval of X-values X > 195 corresponds to the interval of Z-values Z > 0.69.

2. Convert the interval Z > 0.69 to a probability statement.We use the simulation (or some sort of technology) for this step. Below is a picture of the simulation with the settings for this problem. We moved one flag out of the way and the other flag to the position Z = 0.69. For a greater than probability, we want the area to the right of Z = 0.69.

   So we have found that P(X > 195) = P(Z > 0.69) = 0.2451.

Conclusion: This probability is not very low (almost 25%). We conclude that it would be not be surprising to find a randomly chosen individual from this population with a height of 195 cm.

Example

Surprising Heights for Samples of Basketball Players

As before, suppose the heights of individual players are normally distributed with μ = 190 cm and σ = 7.2 cm.

Would it be surprising to find a randomly chosen team of 25 players with a mean height of 195 cm?

We compute the probability that a random sample of 25 players has a mean height of 195 cm or more. We have to look at the distribution of all sample means for samples of size 25. Heres what we know about this sampling distribution:

• The distribution of sample means is normal, even though our sample size is less than 30, because we know the distribution of individual heights is normal. If the individual heights were not normally distributed, we would need a larger sample size before using a normal model for the sampling distribution.
• The mean of the sampling distribution is 195 cm, the same as the mean of the individual heights.
• The standard deviation of the sampling distribution is[latex]\mathrm{σ}\text{}/\sqrt{n}\text{}=\text{}7.2\text{}/\sqrt{25}\text{}=\text{}7.2\text{}/\text{}5\text{}=\text{}1.44[/latex]

Now we can answer this question by computing the probability that a randomly chosen sample of 25 players from this population has mean height greater than 195 cm. To carry out the analysis, lets use [latex]\overline{X}[/latex] to denote the mean height of a random sample of 25 players from this population. Because mean heights are normally distributed, we can convert mean heights to z-scores and use our simulation to find the probability P( [latex]\overline{X}[/latex] > 195).

1. Convert the interval [latex]\overline{X}[/latex] > 195 to an interval of z-scores.Note that the z-score is the number of standard errors the sample mean is from µ. So the z-score calculation for the sampling distribution has mean μ = 190 and standard deviation [latex]\mathrm{σ}\text{}/\sqrt{n}\text{}=\text{}1.44[/latex].The formula for the z-score of [latex]\overline{X}[/latex] is

   [latex]Z\text{}=\text{}\frac{\stackrel{¯}{X}-\mathrm{μ}}{\mathrm{σ}\text{}/\sqrt{n}}[/latex]

   So the z-score of [latex]\overline{X}[/latex] = 195 is

   [latex]\frac{\stackrel{¯}{X}-\mathrm{μ}}{\mathrm{σ}\text{}/\sqrt{n}}\text{}=\text{}\frac{195-190}{1.44}\text{}=\text{}\frac{5}{1.44}\text{}=\text{}3.47[/latex]

   And the interval of [latex]\overline{X}[/latex]-values [latex]\overline{X}[/latex] > 195 corresponds to the interval of Z-values Z > 3.47.

2. Convert the interval Z > 3.47 to a probability statement.Again, we use the simulation as we did in the previous example. Move the left-hand flag out of the way and the right-hand flag to Z = 3.47. For a greater than probability, we want the area to the right of Z = 3.47.

   So we have found that [latex]P(\stackrel{¯}{X}>195)\text{}=\text{}P(Z>3.47)\text{}=\text{}0.0003[/latex].

Conclusion: This probability is very low (much, much less than 1%). We conclude that it would be very surprising to find a random sample of 25 players from this population with a mean height of 195 cm.

Its interesting to notice that the height cutoff we used in these two examples is the same (195 cm). When considering the individual, we concluded that finding a randomly chosen individual with height of 195 cm would not be surprising. However, when we considered the team, we concluded that it would be very surprising to find a random sample of 25 players with a mean height of 195 cm. This makes sense because as sample size grows, variability shrinks (here we considered a sample of size 1 versus a sample of size 25).

Try It

The annual salary of teachers in a certain state X has a mean of μ = $54,000 and standard deviation of σ = $5,000.