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NikAS [45]
3 years ago
11

The life span of a battery is the amount of time the battery will last. The distribution of life span for a certain type of batt

ery is approximately normal with mean 2.5 hours and standard deviation 0.25 hour. Suppose one battery will be selected at random. Which of the following is closest to the probability that the selected battery will have a life span of at most 2.1 hours?
a. 0.055
b. 0.110
c. 0.445
d. 0.890
e. 0.945
Mathematics
1 answer:
Andrew [12]3 years ago
5 0

Answer:

a. 0.055

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 2.5, \sigma = 0.25

Which of the following is closest to the probability that the selected battery will have a life span of at most 2.1 hours?

This is the pvalue of Z when X = 2.1.

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

Z = \frac{2.1 - 2.5}{0.25}

Z = -1.6

Z = -1.6 has a pvalue of 0.055.

So the correct answer is:

a. 0.055

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<em>Hello</em><em>!</em><em>!</em><em>!</em>

<em>Let's</em><em> </em><em>workout</em><em> </em><em>with</em><em> </em><em>this</em><em> </em><em>figure</em><em>.</em>

<em>BC</em><em> </em><em>is</em><em> </em><em>a</em><em> </em><em>chord</em><em>,</em><em> </em><em>O</em><em> </em><em>is</em><em> </em><em>the</em><em> </em><em>centre</em><em> </em><em>and</em><em> </em><em>OA</em><em> </em><em>is</em><em> </em><em>the</em><em> </em><em>perpendicular</em><em> </em><em>bisector</em><em>.</em>

<em>AB</em><em> </em><em>=</em><em> </em><em>1</em><em>/</em><em>2</em><em> </em><em>of</em><em> </em><em>BC</em><em> </em><em> </em><em> </em><em>(</em><em>according</em><em> </em><em>to</em><em> </em><em>circle's</em><em> </em><em>theorem</em><em>)</em>

<em>so</em><em>,</em><em> </em><em>A</em><em> </em><em>B</em><em> </em><em>=</em><em> </em><em>1</em><em>/</em><em>2</em><em> </em><em>×</em><em> </em><em>2</em><em>5</em><em>.</em><em>6</em>

<em>Therefore</em><em>, </em><em> </em><em>the</em><em> </em><em>measure</em><em> </em><em>of</em><em> </em><em>AB</em><em> </em><em>is</em><em> </em><em>12</em><em>.</em><em>8</em><em>.</em>

<em>now</em><em>,</em><em> </em><em>let's</em><em> </em><em>have</em><em> </em><em>a</em><em> </em><em>small</em><em> </em><em>work</em><em> </em><em>with</em><em> </em><em>triangle</em><em> </em><em>AOB</em><em>.</em>

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