Question

Suppose X~ N (12, 2²), then the mean μ = ____; standard deviation σ= ____; Variance σ² =__.

Please get the following probabilities.

1). P(X< 12.6) = P(Z < (12.6-µ)/σ) = P(Z < (12.6 – 12)/2) = P(Z < 0.3) = 0.6179

2). P(X < 10.6) =

3). P(X < 13.8) =

4). P(X < 12) =

5). P(X > 9) = P (Z > (9-µ)/σ) = P(Z > (9-12)/2) = P(Z>-1.5) = 1-P(Z < -1.5) = 1-0.0668 (I got 0.0668 from the z-table) = .9332

6). P(X > 9.4) =

7). P(X > 12) =

8). P(X > 14.2) =

9). P(9
= P( -1.5 < Z < 0.8) = P(Z< 0.8) –P(Z< -1.5) = 0.7881- 0.0668 – from z-table. = 0.7213

10). P (8
11). P (10.6

Answer 1

A normal distribution with mean μ = 12 and standard deviation σ = 2. Probabilities are calculated using z-scores and the standard normal distribution.

Step-by-step explanation:

A normal distribution with mean μ = 12 and standard deviation σ = 2.

To calculate the probabilities:

1. P(X<12.6) = P(Z<0.3) = 0.6179
2. P(X<10.6) = P(Z<-0.7) = 0.2419
3. P(X<13.8) = P(Z<0.9) = 0.8159
4. P(X<12) = P(Z<0) = 0.5
5. P(X>9) = P(Z>-1.5) = 1-0.0668 = 0.9332
6. P(X>9.4) = P(Z>-0.8) = 1-0.2119 = 0.7881
7. P(X>12) = P(Z>0) = 0.5
8. P(X>14.2) = P(Z>1.1) = 1-0.8643 = 0.1357
9. P(9.4
10. P(10.6