Standard Normal Distribution – In Exercises 9 – 13, assume that thermometer read

Question

Standard Normal Distribution – In Exercises 9 – 13, assume that thermometer readings are normally distributed with a mean of 0oC and a standard deviation of 1.00oC. A thermometer is randomly selected and tested, find the probability of each reading. (The given values are in Celsius degrees.) If using technology instead of Table A-2, round answers to four decimal places.

9. Less than 2.33

10. Greater than 1.45

11. Between 0.24 and 2.77

12. Between -2.23 and 1.23

13. Less than -1.65 and Greater than 2.43

Finding Probability - In Exercises 14 and 15, assume that thermometer readings are normally distributed with a mean of 0oC and a standard deviation of 1.00oC. A thermometer is randomly selected and tested, find the indicated probability, where z is the reading in degrees.

14. P(z < 2.50)

15. P(z < -1.88 or z > 1.88)

In Exercises 16 and 17, find the indicated area under the curve of the standard normal distribution, then convert it to a percentage and fill in the blank.

16. About _______% of the area is between z = -0.75 and z = 0.75 (or within 0.75 standard deviations of the mean).

17. About _______% of the area is between z = -1.5 and z = 1.5 (or within 1.5 standard deviations of the mean).

Finding Temperature Values - In Exercise 18, assume that thermometer readings are normally distributed with a mean of 0oC and a standard deviation of 1.00oC. A thermometer is randomly selected and tested, find the temperature reading corresponding to the given information.

18. If 4.0% of the thermometers are rejected because they have readings that are too low and another 4.0% are rejected because they have readings that are too high, find the two readings that are cutoff values separating the rejected thermometers for the others.

Explanation / Answer

Mean ( u ) =0
Standard Deviation ( sd )=1
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a.
P(X < 2.33) = (2.33-0)/1
= 2.33/1= 2.33
= P ( Z <2.33) From Standard Normal Table
= 0.9901                  
b.
P(X > 1.45) = (1.45-0)/1
= 1.45/1 = 1.45
= P ( Z >1.45) From Standard Normal Table
= 0.0735                  
c.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 0.24) = (0.24-0)/1
= 0.24/1 = 0.24
= P ( Z <0.24) From Standard Normal Table
= 0.59483
P(X < 2.77) = (2.77-0)/1
= 2.77/1 = 2.77
= P ( Z <2.77) From Standard Normal Table
= 0.9972
P(0.24 < X < 2.77) = 0.9972-0.59483 = 0.4024                  
d.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < -2.23) = (-2.23-0)/1
= -2.23/1 = -2.23
= P ( Z <-2.23) From Standard Normal Table
= 0.01287
P(X < 1.23) = (1.23-0)/1
= 1.23/1 = 1.23
= P ( Z <1.23) From Standard Normal Table
= 0.89065
P(-2.23 < X < 1.23) = 0.89065-0.01287 = 0.8778                  

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