Aplastic casing for a magnetic disk is composed of two halves. the thickness of each half is normally distributed with a mean of 2 millimeters (mm) and a standard deviation of 0.1 mm, and the halves are independent. determine the mean and s. d. the total thickness of the two halves. what is the probability that the total thickness exceeds 4.3 mm?

The mean is 4 mm and the standard deviation is 0.1414 mm

Step-by-step explanation:

Let's start defining the random variables.

: ''Thickness in mm of the first half''

: ''Thickness in mm of the second half''

The mean for this random variables is 2 (mm)

The standard deviation for this random variables is 0.1 mm

~ N(2,0.1)

~ N(2,0.1)

Now we calculate the variance of each random variable.

Therefore,

The total thickness ''T'' is equal  to

Because of the independence between the halves and that T is the sum between two normally distributed random variables,

T ~ N [μ1 + μ2, ]

Where μ is the mean of the random variable

Where the mean for T is the sum of the means between and and the variance of T is the sum between the variances of and

μ1 + μ2 = 2 + 2 = 4

Var(T1) + Var(T2) = 0.01 + 0.01 = 0.02

The mean for T is 4 mm and the variance is

The standard deviation is

The standard deviation is 0.1414 mm

T ~ N (4,0.1414)

We subtract the mean and then we divide all by the standard deviation to obtained N(0,1)

Z ~ N(0,1) ⇒

1 - φ(2.1216)

Where φ(2.1216) is the value of the cumulative distribution of N(0,1) evaluated in 2.1216 (You can find this value on a table)

step-by-step explanation:

1. the cylinder is a. 54

2.   a.108 /

7. d. 64

8. i. 32

9. c. 125

10. j. none of the above

11. d. 81

12. i. 216

13. e. none of the above

14. g. 240

15. e. none of the above

16. g. 32

17.   e. none of the above