Respuesta :
Given Information:
Mean salary of employees = μ = $5000
Standard deviation = σ = $500
Sample size = n = 40
Required Information:
P(X < $4900) = ?
Answer:
P(X < $4900) = 0.102 = 10.2%
Step-by-step explanation:
We are given a Normal Distribution, which is a continuous probability distribution and is symmetrical around the mean. The shape of this distribution is like a bell curve and most of the data is clustered around the mean. The area under this bell shaped curve represents the probability.
Let X is the random variable that represents the salary of employees at company.
[tex]P(X < \$ 4900) = P(Z < \frac{x - \mu}{\frac{\sigma}{\sqrt{n} } } )\\\\P(X < \$4900) = P(Z < \frac{4900 - 5000}{\frac{500}{\sqrt{40} } } )\\\\P(X < \$4900) = P(Z < \frac{4900 - 5000}{79} )\\\\P(X < \$4900) = P(Z < \frac{-100}{79} )\\\\P(X < \$4900) = P(Z < -1.27)\\[/tex]
The z-score corresponding to -1.27 from the z-table is 0.102
[tex]P(X < \$4900) = 0.102\\P(X < \$4900) = 10.2 \%[/tex]
Therefore, there is 10.2% probability that the salary of a randomly selected employee will be less than $4900.
How to use z-table?
Step 1:
In the z-table, find the two-digit number on the left side corresponding to your z-score.(e.g -1.2, 2.2, etc.)
Step 2:
Then look up at the top of z-table to find the remaining decimal point in the range of 0.00 to 0.09. (e.g. if you are looking for 1.27 then go for 0.07 column)
Step 3:
Finally, find the corresponding probability from the z-table at the intersection of step 1 and step 2.
