Question

The Precision Scientific Instrument Company manufactures thermometers that are supposed to give readings of at the freezing point of water. Tests on a large sample of these thermometers reveal that at the freezing point of water, some give readings below (denoted by negative numbers) and some give readinbove (denoted by positive numbers). Assume that the mean reading is and the standard deviation of the readings is Also assume that the frequency distribution of errors closely resembles the normal distribution. A thermometer is randomly selected and tested. Find the temperature reading corresponding to the given information. Find P 96, the 96th percentile.

Answer 1

Answer:

The $96^{th}$ percentile is 1.751                                    

Step-by-step explanation:

The following information is missing in the question:

Mean,

$\mu = 0^\circ C$

Standard Deviation,

$\sigma = 1^\circ C$

We are given that the distribution of distribution of errors is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

We have to find the value of x such that the probability is 0.96

$P( X < x) = P( z < \displaystyle\frac{x - 0}{1})=0.96$  

Calculation the value from standard normal z table, we have,  

$\displaystyle\frac{x}{1} = 1.751\\\\x = 1.751$

$P_{96} = 1.751$

1.751 degree Celsius is the thermometer reading corresponding to $96^{th}$ percentile.