Question

The age for COVID-19 patients in a country is normally distributed with mean 57.6 years and standard deviation 28 years. A COVID-19 patient was randomly selected from that country. Find the probability that this patient
(i) is below 50 years.
(ii) is between 30 and 75 years.
(iii) 5% of the patients are above k years old. Find k.
(b) The number of COVID-19 patients in 5 different countries are shown in the table below.
Country A B C D E Number of patients 1006 112 1104 926 1852 Test at 10% significance level, if the number of COVID-19 patients is evenly distributed among the five countries.

(c) In 2019, 20% of the students at University X are from China. In a random sample of 500 university students selected recently, it is found that that 130 of them are from China. Test if there is an increase in the percentage of China students in University X at 3% significance level.

Answer 1

(a)

(i)

= 57.6

= 28

To find P(X<50):

Z = (50 - 57.6)/28

= - 0.2714

By Technology, Cumulative Area Under Standard Normal Curve = 0.3930

So,

P(X<50):= 0.3930

So,

Answer is:

0.3930

(ii)

= 57.6

= 28

To find P(30<X<75):

For X = 30:

Z = (30 - 57.6)/28

= - 0.9857

By Technology, Cumulative Area Under Standard Normal Curve = 0.1621

For X = 75:

Z = (75 - 57.6)/28

= 0.6214

By Technology, Cumulative Area Under Standard Normal Curve = 0.7328

So,

P(30<X<75):= 0.7328 - 0.1621 = 0.5707

So,

Answer is:

0.5707

(iii) 5% are above k corresponds to area = 0.50 - 0.05 = 0.45 from mid value to Z on RHS.

Table gives Z = 1.645

So, we get:
Z = 1.645 = (X - 57.6)/28

So,

X = 57.6 + (1.645 X 28)

= 103.66

So,

Answer is:

103.66

(b)

Tst Statistic() is got as follows:

Observed (O)	Expected (E)	(O - E)2/E
1006	5000 X 0.20 = 1000	0.036
112	5000 X 0.20 = 1000	788.544
1106	5000 X 0.20 = 1000	11.236
926	5000 X 0.20 = 1000	5.476
1852	5000 X 0.20 = 1000	725.904
Total = =		1531.196

df = 5 - 1 = 4

= 0.10

From Table, critical value of = 7.7794

Since calculated value of = 1531.196 is greater than critical value of = 7.7794, the difference is significant. Reject null hypothesis.

Conclusion:
The data do not support the claim that the number of COVID-19 patients is evenly distributed among the five countries.

(c)

H0:Null Hypothesis: p 0.20 ( There is not an increase in the percentage of China students in University X )

HA: Alternative Hypothesis: p > 0.20 ( There is an increase in the percentage of China students in University X ) (Claim)

n = 500

= 130/500 = 0.26

= 0.03

From Table, critical valueof Z = 1.88

Test Statistic is given by:

Since calculated value of Z = 3.354 is greater than critical value of Z = 1.88, the difference is significant. reject null hypothesis.

Conclusion :

The data support the claim that there is an increase in the percentage of China students in University X .