Question

Finch beaks One dimension of bird beaks is "depth"-the height of the beak where it arises from the bird's head. During a research study on one island in the Galapagos archipelago, the beak depth of all Medium Ground Finches on the island was found to be Normally distributed with mean $\mathrm{\mu }=9.5\mathrm{millimeters}\left(\text{mm}\right)$and standard deviation $\mathrm{\sigma }=1.0\text{mm}.$

a. Choose an SRS of 5 Medium Ground Finches from this population. Describe the sampling distribution of $\overline{\mathrm{x}}.$

b. Find the probability that $\overline{\mathrm{x}}$estimates $\mathrm{\mu }$within $\pm 0.5\text{mm}$. (This is the probability that $\overline{\mathrm{x}}$takes a value between 9 and 10 mm.

c. Choose an SRS of 50 Medium Ground Finches from this population. Now what is the probability that $\overline{\mathrm{x}}$falls within $\pm 0.05\mathrm{mm}$of $\mathrm{\mu }$? In what sense is the larger sample "better"?

Short answer

(a) The mean 9.5 and standard deviation is 0.4472

(b) The resultant probability is $73.72\%$

(c) The resultant probability is $99.96\%$

Step-by-step solution

Part( a) Step 1: Given information

$\begin{array}{r}\mathrm{\mu }=9.5\\ \mathrm{\sigma }=1.0\\ \mathrm{n}=5\end{array}$

The following formula was used:

${\mathrm{\sigma }}_{\overline{\mathrm{x}}}=\frac{\mathrm{\sigma }}{\sqrt{\mathrm{n}}}$

Part(a) Step 2: Calculation

The sampling distribution of the sample mean is normal because the population distribution is normal $\overline{\mathrm{x}}$This is also typical.

The sample mean has a mean of the sampling distribution.

${\mathrm{\mu }}_{\overline{\mathrm{x}}}=\mathrm{\mu }=9.5$

The standard deviation of the sampling distribution of the sample mean is

${\mathrm{\sigma }}_{\overline{\mathrm{x}}}=\frac{\mathrm{\sigma }}{\sqrt{\mathrm{n}}}=\frac{1.0}{\sqrt{5}}=0.4472$

As a result, the sample mean's sampling distribution is Normal with mean $9.5$and standard deviation$0.4472$

Part(b) step 1: Given information 

$\begin{array}{r}\mathrm{\mu }=9.5\\ \mathrm{\sigma }=1.0\\ \mathrm{n}=5\\ \mathrm{x}=9\text{or}10\end{array}$

The following formula was used:

$\mathrm{z}=\frac{\mathrm{x}-{\mathrm{\mu }}_{\overline{\mathrm{x}}}}{{\mathrm{\sigma }}_{\overline{\mathrm{x}}}}$

Part(b) step 2: Calculation 

The sampling distribution of the sample mean is normal because the population distribution is normal $\overline{\mathrm{x}}$is also typical.

The Z-score is

$\begin{array}{l}\mathbf{z}\mathbf{=}\frac{\mathbf{x}\mathbf{-}{\mathbf{\mu }}_{\overline{\mathbf{x}}}}{{\mathbf{\sigma }}_{\overline{\mathbf{x}}}}\mathbf{=}\frac{\overline{\mathbf{x}}\mathbf{-}\mathbf{\mu }}{\frac{\mathbf{\sigma }}{\sqrt{\mathbf{n}}}}\mathbf{=}\frac{\mathbf{9}\mathbf{-}\mathbf{9}\mathbf{.5}}{\frac{\mathbf{1}\mathbf{.01}}{\sqrt{\mathbf{5}}}}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{.12}\\ \mathbf{z}\mathbf{=}\frac{\mathbf{x}\mathbf{-}{\mathbf{\mu }}_{\overline{\mathbf{x}}}}{{\mathbf{\sigma }}_{\overline{\mathbf{x}}}}\mathbf{=}\frac{\overline{\mathbf{x}}\mathbf{-}\mathbf{\mu }}{\frac{\mathbf{\sigma }}{\sqrt{\mathbf{n}}}}\mathbf{=}\frac{\mathbf{191}\mathbf{-}\mathbf{188}}{\frac{\mathbf{41}}{\sqrt{\mathbf{100}}}}\mathbf{=}\mathbf{-}\mathbf{0}\mathbf{.7}\end{array}$

The normal probability is used to calculate the associating probability.

$P(Z<-1.12)$is given in the first row, beginning with $-1.1$in the column that begins with 0.2 of the normal probability distribution $\mathrm{P}(\mathrm{Z}<1.12)$is given in the first row, beginning with 1.1 in the column that begins with 0.2 of the normal probability distribution

$\begin{array}{c}\mathrm{P}(9<\overline{\mathrm{X}}<10)=\mathrm{P}(-1.12<\mathrm{Z}<1.12)\\ =\mathrm{P}(\mathrm{Z}<1.12)-\mathrm{P}(\mathrm{Z}<-1.12)\\ =0.8686-0.1314=0.7372=73.72\end{array}$

Part(c) step 1: Given information 

$\begin{array}{r}\mathrm{\mu }=9.5\\ \mathrm{\sigma }=1.0\\ \mathrm{n}=5\\ \mathrm{x}=9\text{or}10\end{array}$

The following formula was used:

$\mathrm{z}=\frac{\mathrm{x}-{\mathrm{\mu }}_{\overline{\mathrm{x}}}}{{\mathrm{\sigma }}_{\overline{\mathrm{x}}}}$

Part(c) step 2: calculation

The sampling distribution of the sample mean is normal because the population distribution is normal $\overline{\mathrm{x}}$is also typical.

Z-score is

$\mathrm{z}=\frac{\mathrm{x}-{\mathrm{\mu }}_{\overline{\mathrm{x}}}}{{\mathrm{\sigma }}_{\overline{\mathrm{x}}}}=\frac{\overline{\mathrm{x}}-\mathrm{\mu }}{\frac{\mathrm{\sigma }}{\sqrt{\mathrm{n}}}}=\frac{9-9.5}{\frac{1.0}{\sqrt{50}}}=-3.54$

The normal probability is used to calculate the associating probabilitylocalid="1654674407714" $\mathrm{P}(\mathrm{Z}<-3.54)$is given in the first row, beginning with -3.5 in the column that begins with .04, In the row beginning with, the normal probability table is shown in its most basic form 3.5 and in the column that begins with .04 of the normal probability distribution

$\begin{array}{l}\mathrm{P}(9<\overline{\mathrm{X}}<10)=\mathrm{P}(-3.54<\mathrm{Z}<3.54)\\ =\mathrm{P}(\mathrm{Z}<3.54)-\mathrm{P}(\mathrm{Z}<-3.54)\\ =0.9998-0.0002\\ =0.9996\\ =99.96\%\end{array}$

The larger sample is "better" because the probability of the sample mean being within 0.5 of the population mean is higher with a larger sample, hence our estimations of the population mean are more accurate.