Question

Gestation Periods of Humans. Gestation periods of humans have a mean of $266$days and a standard deviation of $16$days. Let $y$denote the variable "gestation period" for humans.

a. Find the standardized variable corresponding to $y.$

b. What are the mean and standard deviation of the standardized variable?

c. Obtain the z-scores for gestation periods of 227 days and 315 days. Round your answers to two decimal places.

d. Interpret your answers in part (c).

e. Construct a graph similar to Fig. 3.20 on page 146 that shows your results from parts (b) and (c).

Short answer

a) The standardized variable is$z=\frac{y-266}{16}$

b) The mean of the standardized variable is $0.$The standard deviation is $1.$

c) The z-scores for the gestation period of $277,315$days are $-2.44,3.06$respectively.

d) The gestation period is $2.44$standard deviation below the mean and for z-score $3.06$, the gestation period is $3.06$standard deviation above the mean.

e) Graph showing the result:

Step-by-step solution

Step 1/6

Given Information:

The mean of the human gestation period is $266$days.

a) We have to find the standardized variable corresponding to $y$.

$z=\frac{y-\mu }{\sigma }$is the standardized version of $y.$

$z=\frac{y-266}{16}$

Step 2/6

b) We have to find the mean and standard deviation of standardized variable.

Mean:

$\begin{array}{r}{\mu }_z=\frac{\sum z}{N}\\ =\frac{\sum \left(\frac{y-266}{16}\right)}{N}\\ =\frac{\sum (y-266)}{16N}\\ =\frac{\sum y-N\times 266}{16N}\\ =\frac{N\times 266-N\times 266}{16N}\\ =0\end{array}$

Step 3/6

Standard deviation of the standardized variable:

$\begin{array}{r}{\sigma }_z=\sqrt{\frac{\sum {\left(z-{\mu }_z\right)}^2}{N}}\\ =\sqrt{\frac{\sum (z-0{)}^2}{N}}\\ =\sqrt{\frac{\sum z^2}{N}}\\ =1\end{array}$

Therefore mean is zero and S.D is one.

Step 4/6

c) We need to find z-scores.

Z-scores for a gestation period of $227$days:

$\begin{array}{r}z=\frac{y-{\mu }_y}{{\sigma }_y}\\ =\frac{227-226}{16}\\ =-2.4375\\ =-2.44\end{array}$

Z-scores for a gestation period of $315$days:

$\begin{array}{r}z=\frac{y-{\mu }_y}{{\sigma }_y}\\ =\frac{315-226}{16}\\ =3.0625\\ =3.06\end{array}$

Therefore the z-scores are$-2.44,3.06$.

Step 5/6

d) The z-score is $-2.44$, this means that the gestation period is $2.44$standard deviations below the mean.

The z-score is $3.06$, this means that the gestation period is $3.06$standard deviations above the mean.

Step 6/6

e) Graph showing the result:

The graph represents the number of days of gestation and z-scores.