Question

A sample of 9 data yields the following results: sample mean is 4 and sample variance is 3. For unknown reasons, two of the sample values ​​are lost and when redoing the accounts with the remaining data, a new sample mean of 4 is obtained. and a new sample variance of 3/2.

a. Lost data has other values.
b. The missing data is 2 and 6.
c. The missing data is 0 and 8.
d. The missing data is 4 and 4.

A factory produces pistons whose diameters follow a normal distribution with a mean of 50 mm and a standard deviation of 0.01 mm. For a piston to serve, its diameter must be between 49.98 and 50.02 mm. If the diameter is less than 49.98 mm it is rejected; if it is greater than 50.02 mm, it is reprocessed once and the new diameter follows a normal distribution of an average of 49.99 mm, a standard deviation of 0.01 mm.

I. The probability that the piston must be reprocessed is 0.0228.

II. The probability that a reprocessed piston will be rejected is 0.61

a. Only I is correct

b. Only II is correct.

c. None is correct.

d. Both are correct.

Answer 1

Part 1: Since sample variance of 9 data WI (xi – 7)? = 3 i=1 9 i=1 then (ti – 4)? = 8 +3 = 24 Assume 18, Ig are missing. If Ig = 2, Ig = 6 then (xi – 4)2 = 24 – 8 = 16 then new sample variance = 16/6 = 8/3. 7 (r; – 4) TM i=1 Hence option bis not correct. Ś(23 – 4) = 24 – 32 < O hence i=1 7 Again assume Ig = 0, 2g = 8 then Option c. is incorrect. assume Ig = Tg = 4 then (li – 4)2 = 24 then new sample variance 24/6 = 4 hence Option d. is incorrect. Hence correct option a. i=1

Part 2:

Probability that the piston must be reprocessed=P(X>50.02)=0.0228

R code: round(1-pnorm(50.02,50,0.01),4)

The probability that a reprocessed piston will be rejected=P(Y<49.98)=0.1587

R code: round(pnorm(49.98,49.99,0.01),4)

Option: a. Only I is correct