Question

Assume that the readings on the thermometers are normally distributed with a mean of 0 degrees and standard deviation of 1.00degreesC. Assume 2.8​% of the thermometers are rejected because they have readings that are too high and another 2.8​% are rejected because they have readings that are too low. Draw a sketch and find the two readings that are cutoff values separating the rejected thermometers from the others.

Answer 1

Answer:

1.91° and -1.91°.

Step-by-step explanation:

2.8% of the thermometers are rejected on either end of the curve.  The bottom end, where the readings are too far below the mean, will have an area from this point to the left tail of the curve of 0.028.

The top end, where the readings are too far above the mean, will have an area from this point to the left tail of the curve of 1-0.028 = 0.972.

We look in a z table for these values.  We look within the cells of the table; the closest value to 0.028 is 0.0281, which corresponds with a z score of -1.91.  The closest value to 0.972 is 0.9719, which corresponds with a z score of 1.91.

We substitute these values into the z score formula, along with our values for the mean (0) and the standard deviation (1):

$-1.91=\frac{X-0}{1}$

Simplifying the right hand side, X-0 = X; X/1 = X.  This means X = -1.91.

For the second value,

$1.91=\frac{X-0}{1}$

Simplifying the right hand side, X-0 = X; X/1 = X.  This means X = 1.91.

This means the two values are 1.91° and -1.91°.

Answer 2

Using the normal distribution, it is found that the cutoff values are -1.91 ºC and 1.91 ºC.

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

$Z = \frac{X - \mu}{\sigma}$

• 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 $\mu = 0$.
• Standard deviation of 1ºC, thus $\sigma = 1$.

The cutoff values are:

• The bottom 2.8%, thus X when Z has a p-value of 0.028, so X when Z = -1.91.

• The top 2.8%, cutoff by the 100 - 2.8 = 97.2th percentile, which is X when Z has a p-value of 0.972, so X when Z = 1.91.

Then, for Z = -1.91.

$Z = \frac{X - \mu}{\sigma}$

$-1.91 = \frac{X - 0}{1}$

$X = -1.91$

And, the upper cutoff is:

$Z = \frac{X - \mu}{\sigma}$

$1.91 = \frac{X - 0}{1}$

$X = 1.91$

The cutoff values are -1.91 ºC and 1.91 ºC.

The sketch of the situation is given at the end of this question, with the accepted values.

A similar problem is given at https://brainly.com/question/24663213