A random variable X has the following probabilitydistribution: x 4 5 6 P(x) 0.3

Question

A random variable X has the following probabilitydistribution:

x

4                            5                        6

P(x)

0.3                        0.5                      0.2

Find the mean and standard error of the mean for a randomsample of size 2.

b)

It is known that 3% of the persons living in Gujranwala city areknown to have a certain disease. Find the mean and standard errorof sampling distribution of proportion of diseased persons in arandom sample of 500 persons.

x

4                            5                        6

P(x)

0.3                        0.5                      0.2

Explanation / Answer

given tha data, x         4      5      6 p(x)   0.3   0.5      0.2 mean = 4*0.3+5*0.5+6*0.2          =1.2 +2.5 + 1.2         = 4.9 standard deviation = sqrt( x^2*p(x) - (x*p(x))^2)                             =sqrt([16*0.3+25*0.5+36*0.2 ] - [4.9]^2)                            =sqrt(     24.5 - 24.01)                             =sqrt( 0.49)                            =0.7 b)the probability of persons having disease, p =3% =0.03 number of persons, n = 500 mean of sampling distribution of proportion, np= 0.03*500                                                                       = 15 standard error ofproportion,                    SE(p) = sqrt( pq/n)                                                                           =sqrt( 0.03*0.97 / 500)                                                                           =sqrt( 0.0000582)                                                                           = 0.00763

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---