Q8 Let x be a normally distributed random variable with a me

Q8. Let x be a normally distributed random variable with a mean of 10 and a standard deviation of 2. Find the probability that x lies between 11 and 13.6.

A. 0.7266 B.0.2267 C. 0.2654 D. 0.6272 E.0.2726

Q9.Studies shows that gasoline use for compact cars sold in the United States is normally distributed with a mean use of 25.5mpg and a standard deviation of 4.5mpg. If a manufacturer wishes to develop a compact car that outperforms 95% of the current compacts in fuel economy, what must be the gasoline use rate for the new car?

A. 29.3mpg B.32.9mpg C. 95.6mpg D. 65.9mpg E. 56.9mpg

Q10. A producer of soft drinks was fairly certain that her brand of soft drink had a 10% share of the soft drink market. In a market survey involving 2500 consumers of soft drink, letting x be the number of soft drinks purchased calculate the mean and standard deviation by using normal approximation to the binomial distribution.

A. µ=500, =10 B. µ=250, =15 C. µ=300, =10 D. µ=200, =15

E. µ=500, =15

Consider a population N=5 elements whose values are 3, 6,9,12 and 15.If a sample of n=3 is drawn without replacement from N=5, use this information for Q11 and Q12.

Q11. How many possible samples can be obtained?

A. 2 B.10 C. 9 D. 60 E. 56

Q12. If we let x=3, 6,9,12 and 15, what is the probability of selection of each number if all numbers have equal chances of being selected.

A. 0.02 B.0.35 C. 0.2 D. 0.65 E. 0.56

Use the following sampling distribution table for Q13, Q14 and Q15

Sample

Sample values

Sample mean(x_bar)

1

3,6,9

2

3,6,12

7

3

3,6,15

8

4

3,9,12

5

9

6

3,12,15

10

7

6,9,12

9

8

6,9,15

10

9

6,12,15

11

10

9,12,15

12

Q13. Find .

A. 9 B.10 C.6 D.7 E.11

Q14. Find .

A. 10 B.11 C.8 D.11 E.9

Q15. Find .

A. 9,12,15 B. 3,12,15 C. 6,9,12 D.3,9,15

E. 6,12,15

Q16. The central limit theorem states that

A. If random samples of n observations are drawn from a normal population with finite mean µ and standard deviation , then when n is small; the sampling distribution of x_bar is approximately normally distributed.

B. If random samples of n observations are drawn from a normal population with finite mean µ and standard deviation , then when n is small; the sampling distribution of x_bar is approximately binomially distributed.

C. If random samples of n observations are drawn from a non-normal population with finite mean µ and standard deviation , then when n is small; the sampling distribution of x_bar is approximately normally distributed.

D. If random samples of n observations are drawn from a non-normal population with finite mean µ and standard deviation , then when n is large, the sampling distribution of x_bar is approximately normally distributed.

E. If random samples of n observations are drawn from a normal population with finite mean µ and standard deviation , then when n is small; the sampling distribution of x_bar is approximately exponentially distributed.

Q17. The mathematical notation of the central limit theorem is

A. X~N(µx_bar=µ, x_bar=/n^1/2)

B. X~N(µx_bar=µ, x_bar=/n)

C. X~N(µx_bar=µ^1/2, x_bar=/n^1/2)

D. X~N(µx_bar=µ^1/2, x_bar=/n^1/2)

E. X~N(µx_bar=µ, x_bar=^1/2/n)

Use the following information to solve Q18-Q20.

Suppose that you select a random sample of n=25 observations from a population with mean µ=8 and =0.6.then

Q18. Find the mean of x_bar.

A. µx_bar=7.9 B. µx_bar=8.0 C. µx_bar=8.5 D. µx_bar=25

E. µx_bar=6

Q19. Find the standard deviation of x_bar.

A. x_bar=0.22 B. x_bar=0.35 C. x_bar=0.12 D. x_bar=0.25

E. x_bar=0.44

Q20. Find the approximate probability that the sample mean x_bar will be less than 7.9.

A. 0.2663 B.0.2067 C. 0.2654 D. 0.2033 E.0.2726

Sample	Sample values	Sample mean(xbar)
1	3,6,9
2	3,6,12	7
3	3,6,15	8
4	3,9,12
5		9
6	3,12,15	10
7	6,9,12	9
8	6,9,15	10
9	6,12,15	11
10	9,12,15	12

Solution

Q8.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 11) = (11-10)/2
= 1/2 = 0.5
= P ( Z <0.5) From Standard Normal Table
= 0.69146
P(X < 13.6) = (13.6-10)/2
= 3.6/2 = 1.8
= P ( Z <1.8) From Standard Normal Table
= 0.96407
P(11 < X < 13.6) = 0.96407-0.69146 = 0.2726 ~ E.0.2726          

Q9.
P ( Z < x ) = 0.95
Value of z to the cumulative probability of 0.95 from normal table is 1.645
P( x-u/s.d < x - 25.5/4.5 ) = 0.95
That is, ( x - 25.5/4.5 ) = 1.64
--> x = 1.64 * 4.5 + 25.5 = 32.9 ~ B.32.9mpg

Q10.
Mean ( np ) = 250
Standard Deviation ( npq )= 2500*0.1*0.9 = 15
B. µ=250, =15

NOTE: Post the remining in next post