The Precision Scientific Instrument Company manufactures thermometers that are supposed to give readings of 0degreesC 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 0degreesC (denoted by negative numbers) and some give readings above 0degreesC (denoted by positive numbers). Assume that the mean reading is 0degreesC and the standard deviation of the readings is 1.00degreesC. Also assume that the frequency distribution of errors closely resembles the normal distribution. A thermometer is randomly selected and tested. A quality control analyst wants to examine thermometers that give readings in the bottom 4%. Find the temperature reading that separates the bottom 4% from the others. Round to two decimal places. Find P40, the 40th percentile.

a)-0.57 degrees B) 0.57degrees C) 0.25 degrees D) -0.25 degrees

Find Q3, the third quartile.

a) 0.67 degrees B) 0.82 degrees C) -1.3 degrees D) 0.53 degrees

The bottom 4% has a probability of 4% = 4/100 = 0.04. Using a z-table, we find the value closest to 0.04 in the cells of the chart; this is 0.0401. This corresponds to a z value of -1.75.

The formula for a z score is

[tex]z=\frac{X-\mu}{\sigma}[/tex]

Substituting our values for z, the mean and the standard deviation, we have

-1.75 = (X-0)/1

X-0 = X, and X/1 = X; this gives us

-1.75 = X.

The 40th percentile is the value that is greater than 40% of the other data values. This means it has a probability of 0.40. Using a z-table, we find the value closest to 0.40 in the cells of the chart; this is 0.4013. This corresponds to a z value of -0.25.

Substituting this into our formula for a z score along with our values for the mean and the standard deviation,

-0.25 = (X-0)/1

X-0 = X, and X/1 = X; this gives us

-0.25 = X.

Q3 is the same as the 75th percentile. This means it has a probability of 0.75. Using a z-table, we find the value closest to 0.75 in the cells of the chart; this is 0.7486. This corresponds to a z value of 0.67.

Substituting this into our z score formula along with our values for the mean and the standard deviation,

0.67 = (X-0)/1

X-0 = X and X/1 = X; this gives us

0.67 = X.

joaobezerra joaobezerra · Answer 2 · 2021-11-09T16:54:46+01:00

Using the normal distribution, it is found that:

The temperature reading that separates the bottom 4% from the others is -1.75 ºC.

The 40th percentile is of -0.25 ºC, option D.

The third quartile is of 0.67 ºC, option A.

In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

It measures how many standard deviations the measure is from the mean.
After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

In this problem:

Mean of 0ºC, thus [tex]\mu = 0[/tex].
Standard deviation of 1ºC, thus [tex]\sigma = 1[/tex].

The temperature reading that separates the bottom 4% from the others is the 4th percentile, which is X when Z has a p-value of 0.04, so X when Z = -1.75.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]-1.75 = \frac{X - 0}{1}[/tex]

[tex]X = -1.75[/tex]

The temperature reading that separates the bottom 4% from the others is -1.75 ºC.

The 40th percentile is X when Z has a p-value of 0.4, so X when Z = -0.25.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]-0.25 = \frac{X - 0}{1}[/tex]

[tex]X = -0.25[/tex]

The 40th percentile is of -0.25 ºC, option D.

The third quartile is the 75th percentile, as [tex]\frac{3}{4}100 = 75[/tex], which is X when Z has a p-value of 0.75, so X when Z = 0.67.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]0.67 = \frac{X - 0}{1}[/tex]

[tex]X = 0.67[/tex]

The third quartile is of 0.67 ºC, option A.

A similar problem is given at https://brainly.com/question/12403139