Question

The number of texts per day by students in a class is normally distributed with a 

mean of 130 texts and a standard deviation of 20 texts. Identify all true statements.

1. A little more than 2% of the  students text 
under 90 times per day.
2. About 50% of the students  text more than 130 times per  day.
3. About 15% of the students  send over 190  texts per day.
4. About 68% of the students  text between 100 and 130  times.
5. A student texting 130 times  per day would have a z score  of 1.
6. A student texting 160 times  falls between 1 and 2 standard deviations from the mean.

Answer 1

1, 2, 6

Explanation:

The z score shows by how many standard deviations the raw score is above or below the mean. The z score is given by:

`z=(x-\mu)/(\sigma) \\\\where\ x=raw\ score,\mu=mean, \ \sigma=standard\ deviation`

Given that mean (μ) = 130 texts, standard deviation (σ) = 20 texts

1) For x < 90:

`z=(x-\mu)/(\sigma) \\\\z=(90-130)/(20) =-2`

From the normal distribution table, P(x < 90) = P(z < -2) = 0.0228 = 2.28%

Option 1 is correct

2) For x > 130:

`z=(x-\mu)/(\sigma) \\\\z=(130-130)/(20) =0`

From the normal distribution table, P(x > 130) = P(z > 0) = 1 - P(z < 0) = 1 - 0.5 = 50%

Option 2 is correct

3) For x > 190:

`z=(x-\mu)/(\sigma) \\\\z=(190-130)/(20) =3`

From the normal distribution table, P(x > 3) = P(z > 3) = 1 - P(z < 3) = 1 - 0.9987 = 0.0013 = 0.13%

Option 3 is incorrect

4) For x < 130:

`z=(x-\mu)/(\sigma) \\\\z=(130-130)/(20) =0`

For x > 100:

`z=(x-\mu)/(\sigma) \\\\z=(100-130)/(20) =-1.5`

From the normal table, P(100 < x < 130) = P(-1.5 < z < 0) = P(z < 0) - P(z < 1.5) = 0.5 - 0.0668 = 0.9332 = 93.32%

Option 4 is incorrect

5) For x = 130:

`z=(x-\mu)/(\sigma) \\\\z=(130-130)/(20) =0`

Option 5 is incorrect

6) For x = 130:

`z=(x-\mu)/(\sigma) \\\\z=(160-130)/(20) =1.5`

Since 1.5 is between 1 and 2, option 6 is correct