Question

Assume that the readings at freezing on a batch of thermometers are normally distributed with a mean of 0°C and a standard deviation of 1.00°C. A single thermometer is randomly selected and tested. Find P36, the 36-percentile. This is the temperature reading separating the bottom 36% from the top 64%.

P_36 = ?
°C

Answer 1

Answer:

Use z = (x - μ)/σ or  x = zσ + μ

Given: μ = 0 oC, σ = 1.00 oC

94% = .9400 on a z-table falls half-way between z = 1.55 and z = 1.56 --> z = 1.555

x = 1.555(1.00 oC) + 0 oC = 1.555 oC

P94 = 1.555 oC

Step-by-step explanation: I had this on my quiz and hope, it helps if i am wrong please tell me!

Answer 2

The value of P36 is 0.14058°C

To calculate the P36:

Use z = (x - μ)/σ

Given: μ = 0 °C, σ = 1.00 °C

36% = .3600 on a z-table falls half-way between z = 0.64058 and z = 0.35942 --> z = 0.14058

x = 0.14058(1.00 °C) + 0 °C = 0.14058 °C

P36 = 0.14058 °C

Hence the value of P36 is 0.14058.

To learn more about how to calculate the freezing point of the bath refer:https://brainly.com/question/12940710

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