SOLUTION: I need help with this question. Question: The temperature recorded by a certain thermometer when placed in boiling water (true temperature 100 degree C) is normally distributed

Question 130977: I need help with this question.
Question: The temperature recorded by a certain thermometer when placed in boiling water (true temperature 100 degree C) is normally distributed with mean μ = 99.8 degree C and standard deviation σ = 0.1 degree C. [Please draw a density curve before you do any calculations.]
1). What is the probability that the thermometer reading is greater than 100 degree C?
2). What is the probability that the thermometer reading is exactly 100 degree C?
3). What is the probability that the thermometer reading is within � 0.05 degree C?

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
The temperature recorded by a certain thermometer when placed in boiling water (true temperature 100 degree C) is normally distributed with mean μ = 99.8 degree C and standard deviation σ = 0.1 degree C. [Please draw a density curve before you do any calculations.]
------------
Draw a normal curve; put mean = 99.8.
Calculate the z-value of the termperature in each of the problems using
z(x) = (x-mu)/sigma
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1). What is the probability that the thermometer reading is greater than 100 degree C?
z(100) = (100-99.8)/0.1 = 0.02/0.1 = 0.2
P(z>0.2) = 0.4207
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2). What is the probability that the thermometer reading is exactly 100 degree C?
The probability of any particular value is zero
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3). What is the probability that the thermometer reading is within � 0.05 degree C?
I'll assume you mean within 0.05 degrees of the mean.
Find the z-value of 99.8+0.05= 99.85
z(99.85) = (99.85-99.8)/0.1 = 0.5
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P(99.8-0.05 < temp < 99.5+0.05) = P(-0.5 < z < 0.5) = 0.3829
============
Cheers,
Stan H.

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