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
and standard deviation
, the z-score of a measure X is given by:

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 
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



has a p-value of 0.0455
X = -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:
I think the answer is d or b if im wrong sorry hope this helped tho :p
Answer:
1/9
Step-by-step explanation:
3 red
2 blue
5 green
Total
3
+
2
+
5
=10
Probability of blue is
2
10
~~~~~~~~~~~~~~~~~~~~~~~
Second condition:
3 red
1 blue
5 green
New total
3
+
1
+
5
=
9
Probability of green is
5
9
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Total probability of selecting this path is:
2
10
×
5
9
=
2
9
×
5
10
=
2
9
×
1
2
=
1
9
1/4*x=3/8
1*x=3*4/8
X=12/8
3/2