Pharmacy exam 2020-10-26

Degrees: Pharmacy, Biotechnology
Date: October 26, 2020

Question 1

The table below shows the daily number of patients hospitalized in a hospital during the month of September.

$\begin{array}{cr} Patients & Frequency \\ (10, 14] & 6 \\ (14, 18] & 10 \\ (18, 22] & 7 \\ (22, 26] & 6 \\ (26, 30] & 1 \end{array}$

Study the spread of the 50% of central data.
Compute the mean and study the dispersion with respect to it.
Study the normality of the patients distribution.
If the mean was 35 patients and the variance 40 patients $^{2}$ during the month of April, which month had a higher relative variability?
Which number of people hospitalized was greater, 20 persons in September or 40 in April?

Use the following sums for the computations:
$\sum x_{i} n_{i} = 544$ patients, $\sum x_{i}^{2} n_{i} = 10464$ patients $^{2}$ , $\sum (x_{i} - \bar{x})^{3} n_{i} = 736.14$ patients $^{3}$ and $\sum (x_{i} - \bar{x})^{4} n_{i} = 25367.44$ patients $^{4}$ .

Solution

$Q_{1} = 16$ patients, $Q_{3} = 20$ patients and $I Q R = 4$ patients. Thus the central dispersion is small.
$\bar{x} = 18.1333$ patients, $s^{2} = 19.9822$ patients $^{2}$ , $s = 4.4701$ patients and $c v = 0.2465$ . Thus, the dispersion with respect to the mean is small and the mean represents well.
$g_{1} = 0.2747$ and $g_{2} = - 0.8823$ . As the coefficient of skewness and the coefficient of kurtosis fall between -2 and 2, we can assume that the sample comes from a normal population.
Let $Y$ be the daily number of patients hospitalized during April. Then, $c v_{y} = 0.1807$ . Since the coefficient of variation in September is greater than the one in April, there is a relative higher variability in September.
September: $z (20) = 0.4176$ .
April: $z (40) = 0.7906$ .
Thus, 40 patients hospitalized in April is relatively higher than 20 in September as its standard score is greater.

Question 2

The chart below shows the distribution of scores in three subjects.

Which subject is more difficult?
Which subject has more central dispersion?
Which subjects have outliers?
Which subject is more asymmetric?

Solution

Subject $Y$ because its scores are smaller.
Subject $X$ because the box is wider.
Subject $Z$ because there is a score out of the whiskers.
Subject $Z$ because the distance from the first quartile to the median (left side of the box) is greater than the distance from the third quartile to the median (right side of the box).

Question 3

In a sample of 10 families with a son older than 20 it has been measured the height of the father ( $X$ ), the mother ( $Y$ ) and the son ( $Z$ ) in centimetres, getting the following results:

$\sum x_{i} = 1774$ cm, $\sum y_{i} = 1630$ cm, $\sum z_{i} = 1795$ cm,
$\sum x_{i}^{2} = 315300$ cm $^{2}$ , $\sum y_{i}^{2} = 266150$ cm $^{2}$ , $\sum z_{i}^{2} = 322737$ cm $^{2}$ ,
$\sum x_{i} y_{j} = 289364$ cm $^{2}$ , $\sum x_{i} z_{j} = 318958$ cm $^{2}$ , $\sum y_{i} z_{j} = 292757$ cm $^{2}$ .

On which height does the height of the son depend more linearly, the height of the father or the mother?
Using the best linear regression model, predict the height of a son with a father 181 cm tall and a mother 163 cm tall.
According to the linear model, how much will increase the height of the son for each centimetre that increases the height of the father? And for each centimetre that increases the height of the mother?
How would the reliability of the prediction be if the heights were measured in inches? (An inch is 2.54 cm).

Solution

$\bar{x} = 177.4$ cm, $s_{x}^{2} = 59.24$ cm $^{2}$ ,
$\bar{y} = 163$ cm, $s_{y}^{2} = 46$ cm $^{2}$ ,
$\bar{z} = 179.5$ cm, $s_{z}^{2} = 53.45$ cm $^{2}$ ,
$s_{x z} = 52.5$ cm $^{2}$ and $s_{y z} = 17.2$ cm $^{2}$ .
$r_{x z}^{2} = 0.8705$ and $r_{y z}^{2} = 0.1203$ , thus the height of the son depends linearly more on the height of the father since the $r_{x z}^{2} > r_{y z}^{2}$ .
Regression line of $Z$ on $X$ : $z = 22.2836 + 0.8862 x$ and $z (181) = 182.6904$ cm.
The height of the son will increase $0.8862$ cm per cm of the height of the father and $0.3739$ cm per cm of the height of the mother.
The reliability of the prediction will be the same, as after applying the same linear transformation to $X$ and $Z$ , the variances are multiplied by the square of the slope and the covariance is also multiplied by the square of the slope.

Last updated on Jan 23, 2021