Respuesta :
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([tex]\mu=1,000 ,\sigma^{2} = 100^{2}[/tex])
The z-score probability distribution for a normal distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = mean income = $1,000
[tex]\sigma[/tex] = 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 = [tex]\frac{X-\mu}{\sigma}[/tex] = [tex]\frac{1,100-1,000}{100}[/tex] = 1
Hence, the z-score (value of z) for an income of $1,100 is 1.
