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levacccp [35]
2 years ago
15

The average yearly snowfall in Chillyville is Normally distributed with a mean of 55 inches. If the snowfall

Advanced Placement (AP)
1 answer:
GuDViN [60]2 years ago
7 0

Answer:

The standard deviation is 4.83 inches.

Explanation:

We are given that the average yearly snowfall in Chillyville is Normally distributed with a mean of 55 inches.  

The snowfall  in Chillyville exceeds 60 inches in 15% of the years, we have to find the standard deviation.

Let X = <u><em>the average yearly snowfall in Chillyville</em></u>.

The z-score probability distribution for the normal distribution is given by;

                                 Z = \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = mean amount of rainfall = 55 inches

            \sigma = standard deviation  

Now, it is stated that the snowfall  in Chillyville exceeds 60 inches in 15% of the years, that means;

        P(X > 60 inches) = 0.15

        P( \frac{X-\mu}{\sigma} > \frac{60-55}{\sigma} ) = 0.15

        P(Z > \frac{60-55}{\sigma} ) = 0.15

In the z table, the critical value of z that represents the top 15% of the area is given as  1.0364, that means;

                       \frac{60-55}{\sigma} = 1.0364

                       \sigma} = \frac{5}{1.0364} = 4.83 inches

Hence, the standard deviation is 4.83 inches.

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