17. BATTERY The life of a certain brand of AA battery is normally distributed with u = 8 hours and o = 1.5 hours. Find each probability. (Example 5) a. The battery will last less than 6 hours. b. The battery will last more than 12 hours. c. The battery will last between 8 and 9 hours.

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Solve the Battery Question as the same steps in the image and give the answer of probability as a precentage please. Thank you in advance, I appreciate your compassion.
17. BATTERY The life of a certain brand of AA battery
is normally distributed with u = 8 hours and
o = 1.5 hours. Find each probability. (Example 5)
a. The battery will last less than 6 hours.
b. The battery will last more than 12 hours.
c. The battery will last between 8 and 9 hours.
Transcribed Image Text:17. BATTERY The life of a certain brand of AA battery is normally distributed with u = 8 hours and o = 1.5 hours. Find each probability. (Example 5) a. The battery will last less than 6 hours. b. The battery will last more than 12 hours. c. The battery will last between 8 and 9 hours.
Example 5 Find Probabilities
METEOROLOGY The temperatures for one month for a city in California are normally distributed
with u = 81° and o = 6°. Find each probability, and use a graphing calculator to sketch the
corresponding area under the curve.
a. P(70° < X< 90°)
The question is asking for the percentage of temperatures that were between 70' and 90°. First,
find the corresponding z-values for X = 70 and X = 90.
Formula for z-values
70 – 81
X=70, µ= 81, and a = 6
= -1.83
Simplify.
Use 90 to find the other z-value.
X - u
Formula for z-values
123
X = 90, µ=81, and o= 6
90 – 81
= 1.5
Simplify.
You can use a graphing calculator to display the area
that corresponds to any z-value by selecting 2nd [DISTR].
Then, under the DRAW menu, select ShadeNorm
(lower z value, upper z value). The area between
z = -1.83 and z = 1.5 is 0.899568, as shown.
Area=.899568
1ow= "1.83UP=1.5.
Therefore, approximately 90% of the temperatures were
between 70 and 90.
(-4, 4] scl: 1 by (0, 0.5] scl: 0.125
Transcribed Image Text:Example 5 Find Probabilities METEOROLOGY The temperatures for one month for a city in California are normally distributed with u = 81° and o = 6°. Find each probability, and use a graphing calculator to sketch the corresponding area under the curve. a. P(70° < X< 90°) The question is asking for the percentage of temperatures that were between 70' and 90°. First, find the corresponding z-values for X = 70 and X = 90. Formula for z-values 70 – 81 X=70, µ= 81, and a = 6 = -1.83 Simplify. Use 90 to find the other z-value. X - u Formula for z-values 123 X = 90, µ=81, and o= 6 90 – 81 = 1.5 Simplify. You can use a graphing calculator to display the area that corresponds to any z-value by selecting 2nd [DISTR]. Then, under the DRAW menu, select ShadeNorm (lower z value, upper z value). The area between z = -1.83 and z = 1.5 is 0.899568, as shown. Area=.899568 1ow= "1.83UP=1.5. Therefore, approximately 90% of the temperatures were between 70 and 90. (-4, 4] scl: 1 by (0, 0.5] scl: 0.125
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