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VikaD [51]
3 years ago
13

The mean of a normally distributed group of weekly incomes of a large group of executives is $1,000 and the standard deviation i

s $100. What is the z-score (value of z) for an income of $1,100
Mathematics
1 answer:
olga_2 [115]3 years ago
4 0

Answer:

The z-score (value of z) for an income of $1,100 is 1.

Step-by-step explanation:

We are given that the mean of a normally distributed group of weekly incomes of a large group of executives is $1,000 and the standard deviation is $100.

<em>Let X = group of weekly incomes of a large group of executives</em>

So, X ~ N(\mu=1,000 ,\sigma^{2}  = 100^{2})

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

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

where, \mu = mean income = $1,000

            \sigma = standard deviation = $100

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) 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.

Now, we are given an income of $1,100 for which we have to find the z-score (value of z);

So, <em><u>z-score</u></em> is given by = \frac{X-\mu}{\sigma} = \frac{1,100-1,000}{100} = 1

<em>Hence, the z-score (value of z) for an income of $1,100 is 1.</em>

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