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Answer :
Probability that Thermometer Reading is less than 99.5°C:
We are given that temperature recorded by a certain Thermometer when placed in boiling water (the true temperature is 100 °C) is normally distributed with mean µ = 99.8°C and Standard Deviation σ = 0.1°C.
We need to find the probability that thermometer reading is less than 99.5 °C.
Using the z-value formula: z = (x-µ)/σFor x = 99.5°C, µ = 99.8°C, and σ = 0.1°C,z = (99.5 - 99.8) / 0.1= -3So, P(x < 99.5) = P(z < -3) = 0.0013 [using normal distribution table]
Probability that thermometer reading is greater than 100°C:
Using the z-value formula: z = (x-µ)/σFor x = 100°C, µ = 99.8°C, and σ = 0.1°C,z = (100 - 99.8) / 0.1= 2
So, P(x > 100) = P(z > 2) = 0.0228 [using normal distribution table]
Probability that thermometer reading is within ± 0.05 °C of the true temperature of 100 °C:
Using the z-value formula: z1 = (x1-µ)/σ, z2 = (x2-µ)/σFor x1 = 99.95°C, x2 = 100.05°C, µ = 100°C, and σ = 0.1°C,z1 = (99.95 - 100) / 0.1= -0.5, z2 = (100.05 - 100) / 0.1= 0.5P(99.95 < x < 100.05) = P(-0.5 < z < 0.5) = 0.3829 [using normal distribution table]
Third quartile of the recorded temperature values:
Using the z-value formula: z = (x-µ)/σ
For third quartile, 75th percentile is used
. As we know that 75% of the area is below
So the remaining 25% will be above it.
Using the Normal Distribution Table, the z-value corresponding to 0.25 probability is 0.67.
For µ = 99.8°C, σ = 0.1°C and z = 0.67,
we get, x = µ + zσ= 99.8 + 0.67(0.1)= 99.8665 °C
Therefore, the third quartile of the recorded temperature values is 99.8665°C.
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