Well lets see A, B and D are too large in the parenthesis that the numbers outside of the parenthesis will use to multiply considering 42 + 30.
6(7 + 5)
6(12)
72

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3 years ago

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

olga_2 [115]

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.

Let X = group of weekly incomes of a large group of executives

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, z-score is given by = $\frac{X-\mu}{\sigma}$ = $\frac{1,100-1,000}{100}$ = 1