Can you please help me

Assume the readings on thermometers are normally distributed with a mean of 0°C and a standard deviation of 1.00°C. Find the pro

lisov135 [29]

Answer:

0.0326 = 3.26% probability that a randomly selected thermometer reads between −2.23 and −1.69.

The sketch is drawn at the end.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

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

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

The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Mean of 0°C and a standard deviation of 1.00°C.

This means that $\mu = 0, \sigma = 1$

Find the probability that a randomly selected thermometer reads between −2.23 and −1.69

This is the p-value of Z when X = -1.69 subtracted by the p-value of Z when X = -2.23.

X = -1.69

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

$Z = \frac{-1.69 - 0}{1}$

$Z = -1.69$

$Z = -1.69$ has a p-value of 0.0455

X = -2.23

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

$Z = \frac{-2.23 - 0}{1}$

$Z = -2.23$

$Z = -2.23$ has a p-value of 0.0129

0.0455 - 0.0129 = 0.0326

0.0326 = 3.26% probability that a randomly selected thermometer reads between −2.23 and −1.69.

Sketch:

3 0

3 years ago

In this exercise, T: R2 → R2 is a function. For each of the following parts, state why T is not linear. (a) T(a1, a2)= (1, a2) (

DENIUS [597]

Answer:

a) T is not homogeneous
b) T is not additive
c) T is not homogeneous
d) T is not additive
e) T is not additive

Step-by-step explanation:

In each case there is an example where one property of linear functions fails.

a) $T(2(1,0))=T((2,0))=(1,0); 2T((1,0))=2(1,0)=(2,0)$ . These vectors are not equal, then T doesn't satisfy the condition of scalar multiplication (homogeneity).

b) $T((1,2)+(2,3))=T(3,5)=(3,9); T((1,2))+T((2,3))=(1,1)+(2,4)=(3,5)$ . Because these vectors are not equal, T doesn't satisfy the property of vector addition (additivity).