Physiotherapy exam 2017-05-19

Degrees: Physiotheraphy
Date: May 19, 2017

Probability and random variables

Question 1

The prevalence of sciatica in a population is 3%. The Lasegue’s test is a neurotension test that is used to diagnose the sciatica with a sensitivity of 91% and a specificity of 26%. On the other hand, there is an alternative test with a sensitivity of 80% and a specificity of 90%.

Compute the positive predictive value for the Lasegue’s test.
Assuming that the tests are independent, compute the probability of having a positive outcome in both tests.
Compute the probability of getting a wrong diagnose in the Lesegue’s test or in the alternative test.
Which test is better as a screening test (to rule out the sciatica)?

Solution

$P P V = P (D | +) = 0.0366$ . It is not a goot test to confirm the sciatica as the post test probability of having the sciatica for a positive outcome is very low.
Naming $L ⁺$ to the event of having a positive outcome in Lasegue’s test and $A ⁺$ to the event of having a positive outcome in the alternative test: $P (L^{+} \cap A^{+}) = P (L^{+}) P (A^{+}) = 0.7451 \cdot 0.121 = 0.0902$ .
Naming $W L$ to the event of having a wrong diagnose with Lasegue’s test and $W A$ to the event of having a wrong diagnose with the alternative test: $P (W L \cup W A) = P (W L) + P (W A) - P (W L \cap W A) = 0.7205 + 0.103 - 0.7205 \cdot 0.103 = 0.7493$ .
Lesegue test: $N P V = P (\overset{―}{D} | -) = 0.9894$ . Alternative test: $N P V = P (\overset{―}{D} | -) = 0.9932$ . Thus, the alternative test is better to rule out the sciatica.

Question 2

A physiotherapist opens a clinic and use the social networks to advertise it. In particular he send a friend request to 20 contacts on Facebook. If the probability that a Facebook user accept the friend request is 80%, what is the probability that more than 18 accept the friend request? What is the expected number of friend requests accepted?

Solution

Naming

X

to the number of accepted friend request,

X \sim B (20, 0.8)

and

P (X > 18) = 0.0692

. The expected number of accepted friend request is

16

Question 3

According to a study of the Information Society of Spain in 2013, the spanish checks the mobile phone 150 times a day in average. What is the probability that a spanish person checks the mobile phone more than 2 times an hour?

Solution

Naming $X$ to the number of times that a spanish person checks the phone in an hour, $X \sim P (6.25)$ and $P (X > 2) = 0.9483$ .

Question 4

The the cervical rotation in a population follows a normal probability distribution model with mean 58º and standard deviation 6º.

Between what values are the cervical rotation of the central 50% of the population?
Taking into account the precision of the measurement instrument, a goniometer, a rotation less than 53º is considered a mobility limitation. If we take a random sample of 100 persons from this population, what is the expected number of persons with mobility limitation in the sample?

Solution

Naming $X$ to the cervical rotation, $X \sim N (58, 6)$ .

$(Q 1, Q 3) = (53.9531, 62.0469)$ .
$P (X < 53) = 0.2023$ and the expected number of persons with mobility limitation in a sample of 100 persons is $20.2328$ .

Last updated on Sep 15, 2020