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:
[tex]z=\frac{x-\mu}{\sigma} \\\\where\ x=raw\ score,\mu=mean, \ \sigma=standard\ deviation[/tex]
Given that mean (μ) = 130 texts, standard deviation (σ) = 20 texts
1) For x < 90:
[tex]z=\frac{x-\mu}{\sigma} \\\\z=\frac{90-130}{20} =-2[/tex]
From the normal distribution table, P(x < 90) = P(z < -2) = 0.0228 = 2.28%
Option 1 is correct
2) For x > 130:
[tex]z=\frac{x-\mu}{\sigma} \\\\z=\frac{130-130}{20} =0[/tex]
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:
[tex]z=\frac{x-\mu}{\sigma} \\\\z=\frac{190-130}{20} =3[/tex]
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:
[tex]z=\frac{x-\mu}{\sigma} \\\\z=\frac{130-130}{20} =0[/tex]
For x > 100:
[tex]z=\frac{x-\mu}{\sigma} \\\\z=\frac{100-130}{20} =-1.5[/tex]
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:
[tex]z=\frac{x-\mu}{\sigma} \\\\z=\frac{130-130}{20} =0[/tex]
Option 5 is incorrect
6) For x = 130:
[tex]z=\frac{x-\mu}{\sigma} \\\\z=\frac{160-130}{20} =1.5[/tex]
Since 1.5 is between 1 and 2, option 6 is correct
