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Vilka [71]
2 years ago
6

The table shows the precipitation in December, in inches, for Washington state from 2005 to 2013. What was the mean amount of pr

ecipitation during this time? Round to the nearest hundredth.
A) 1.97 Inches
B) 2.61 Inches
C) 5.33 Inches
D) 5.98 Inches

Mathematics
2 answers:
polet [3.4K]2 years ago
7 0
Hello,
Please, see the attached file.
Thanks.

zhannawk [14.2K]2 years ago
5 0

Answer : C) 5.33 Inches

The table shows the precipitation in December, in inches, for Washington state from 2005 to 2013.

We are given with 9 values.

number of values =9

To find the mean amount of precipitation during this time we add all the precipitation then divide by number of values

Mean amount of precipitation = \frac{Sum of precipitation}{number of values}

Mean amount of precipitation = \frac{5.98+6.05+7.59 +5.27+ 3.11+7.18+ 2.95 + 7.22+2.61}{9}

= \frac{47.98}{9} = 5.32888

The mean amount of precipitation = 5.33 inches

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A random variable x follows a normal distribution with mean d and standard deviation o=2. It is known that x is less than 5 abou
Vaselesa [24]

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The mean of this distribution is approximately 3.96.

Step-by-step explanation:

Here's how to solve this problem using a normal distribution table.

Let z be the

\displaystyle z = \frac{x - \mu}{\sigma}.

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\displaystyle z = \frac{5 - \mu}{2}.

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The problem stated that P(X \le 5) = 69.85\% = 0.6985. Hence, P(Z \le z) = 0.6985.

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Since P(Z \le z) = 0.6985 and is greater than P(Z \le 0) = 0.50, z > 0. As a result, P(Z \le z) can be written as the sum of P(Z < 0) and P(0 \le Z \le z). Besides, P(Z < 0) = P(Z \le 0) = 0.50. As a result:

\begin{aligned}&P(Z \le z)\\ &= P(Z < 0) + P(0 \le Z \le z) \\ &= 0.50 + P(0 \le Z \le z)\end{aligned}.

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Lookup 0.1985 on a normal distribution table. The corresponding z-score is 0.52. (In other words, P(0 \le Z \le 0.52) = 0.1985.)

Given that

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Solve the equation \displaystyle z = \frac{x - \mu}{\sigma} for the mean, \mu:

\displaystyle 0.52 = \frac{5 - \mu}{2}.

\mu = 5 - 2 \times 0.52 = 3.96.

3 0
3 years ago
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