A company manufactures a part for which there is a desired weight. There is a tolerance of N%, meaning that the range between minus and plus N% of the desired weight is acceptable. The tolerance is 1.5% for this particular part. Given the following vector W where the desired weight is 97. [ 97 98.2 95.5 -97.2 96.2 100 99.9 99.8 96.1 ] Determine what the tolerance is for 97 and place the values in variables low_97 and high_97. Determine if each of the values in vector W is within the tolerance or not. It is acceptable to just have a logical vector called within with 0s and 1s as output. Create an output matrix mat where the first row is W and the second row is whether or not W is within the tolerance.

Step-by-step explanation:

(A) Determine what the tolerance is for the desired weight 97

The tolerance N% is equal to 1.5% of 97

1.5/100 × 97 = 1.455

Lower limit = 97 - 1.455 = 95.545

Upper limit = 97 + 1.455 = 98.455

(B) Place the vector values into two groups;

- Variables lower than 97

95.5, -97.2, 96.2, 96.1

- Variables higher than 97

98.2, 100, 99.9, 99.8

(C) Determine if each of the values is within the tolerance bracket or not

1. Value 97 is within the tolerance. As a matter of fact, it is the desired weight.

2. Value 98.2 is within the tolerance

3. Value 95.5 is within the tolerance

4. Value -97.2 is far below the tolerance bracket or lower limit

5. Value 96.2 is within the tolerance

6. Value 100 is outside the tolerance

7. Value 99.9 is outside the tolerance

8. Value 99.8 is outside the tolerance

9. Value 96.1 is within the tolerance

step-by-step explanation:

y = | -2(-2) |     →     y = | 4 |     →     y = 4       ⇒   (-2, 4)

y = | -2(-1) |     →     y = | 2 |     →       y =   2       ⇒   (-1, 2)    

y = | -2(0) |     →     y = | 0 |     →       y = 0         ⇒   (0, 0)

y = | -2(1) |   c →     y = | -2 |     →     y =   2         ⇒   (1, 2)    

y = | -2(2) |     →     y = | - 4 |     →     y =   4         ⇒   (2, 4)  

hello,

b) 62.47% of the students are between 14 and 18 years old.

step-by-step explanation:

mean = 15 and standard deviation = 2

student between 14 to 18

formula used z-score

for age 14 , z score

using z-table , p(z> -0.5)=0.3085

for age 18 , z score

using z-table , p(z< 1.5)=0.9332

p(-0.5< z< 1.5)=0.9332-0.3085 = 0.6247

thus, 62.47% of the students are between 14 and 18 years old.

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