Question

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

Answer 1

Answer:

1, 2, 6

Step-by-step 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:

$z=\frac{x-\mu}{\sigma} \\\\where\ x=raw\ score,\mu=mean, \ \sigma=standard\ deviation$

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

1) For x < 90:

$z=\frac{x-\mu}{\sigma} \\\\z=\frac{90-130}{20} =-2$

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

Option 1 is correct

2) For x > 130:

$z=\frac{x-\mu}{\sigma} \\\\z=\frac{130-130}{20} =0$

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:

$z=\frac{x-\mu}{\sigma} \\\\z=\frac{190-130}{20} =3$

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:

$z=\frac{x-\mu}{\sigma} \\\\z=\frac{130-130}{20} =0$

For x > 100:

$z=\frac{x-\mu}{\sigma} \\\\z=\frac{100-130}{20} =-1.5$

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:

$z=\frac{x-\mu}{\sigma} \\\\z=\frac{130-130}{20} =0$

Option 5 is incorrect

6)  For x = 130:

$z=\frac{x-\mu}{\sigma} \\\\z=\frac{160-130}{20} =1.5$

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