Question

Q) 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

Q.

the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/ 2σ^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 57.6
standard Deviation ( sd )= 28
a.
i.
probability that this patient is below 50 years
P(X < 50) = (50-57.6)/28
= -7.6/28= -0.2714
= P ( Z <-0.2714) From Standard Normal Table
= 0.393
ii.
probability that this patient is between 30 and 75 years.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 30) = (30-57.6)/28
= -27.6/28 = -0.9857
= P ( Z <-0.9857) From Standard Normal Table
= 0.1621
P(X < 75) = (75-57.6)/28
= 17.4/28 = 0.6214
= P ( Z <0.6214) From Standard Normal Table
= 0.7328
P(30 < X < 75) = 0.7328-0.1621 = 0.5707
iii.
P ( Z > x ) = 0.05
Value of z to the cumulative probability of 0.05 from normal table is 1.6449
P( x-u / (s.d) > x - 57.6/28) = 0.05
That is, ( x - 57.6/28) = 1.6449
--> x = 1.6449 * 28+57.6 = 103.6559
k value is 103.6559
b.
Assumed data,
population mean(u)=940 because not given in the data
Given that,
sample mean, x =1000
standard deviation, s =618.6227
number (n)=5
null, Ho: μ=940
alternate, H1: μ!=940
level of significance, α = 0.1
from standard normal table, two tailed t α/2 =2.132
since our test is two-tailed
reject Ho, if to < -2.132 OR if to > 2.132
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =1000-940/(618.6227/sqrt(5))
to =0.2169
| to | =0.2169
critical value
the value of |t α| with n-1 = 4 d.f is 2.132
we got |to| =0.2169 & | t α | =2.132
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 0.2169 ) = 0.8389
hence value of p0.1 < 0.8389,here we do not reject Ho
ANSWERS
---------------
null, Ho: μ=940
alternate, H1: μ!=940
test statistic: 0.2169
critical value: -2.132 , 2.132
decision: do not reject Ho
p-value: 0.8389
we do not have enough evidence to support the claim that population mean is 940.
c.
Given that,
possibile chances (x)=130
sample size(n)=500
success rate ( p )= x/n = 0.26
success probability,( po )=0.2
failure probability,( qo) = 0.8
null, Ho:p=0.2
alternate, H1: p>0.2
level of significance, α = 0.03
from standard normal table,right tailed z α/2 =1.881
since our test is right-tailed
reject Ho, if zo > 1.881
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.26-0.2/(sqrt(0.16)/500)
zo =3.3541
| zo | =3.3541
critical value
the value of |z α| at los 0.03% is 1.881
we got |zo| =3.354 & | z α | =1.881
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value: right tail - Ha : ( p > 3.3541 ) = 0.0004
hence value of p0.03 > 0.0004,here we reject Ho
ANSWERS
---------------
null, Ho:p=0.2
alternate, H1: p>0.2
test statistic: 3.3541
critical value: 1.881
decision: reject Ho
p-value: 0.0004
we have enough evidence to support the claim that if there is an increase in the percentage of China students in University X