Answer:
Explanation:
The first part of the question, with the input data, is missing.
This is missing part:
- We assume that X, the initial salary offer for education majors is normally distributed with a mean of $46,292 and a standard deviation of $4,320
Solution
In the first question: The probability that a randomly selected education major had a starting salary offer exceeding $52,350 is ________
We need to calculate the Z-score.
The Z-score reflects the standardized value of the random variable, indicating how many standard deviations a value is from the mean:



The standard normal distribution tables provide cumulative probabilities as the area under the bell-shaped curve. P(Z>1.40) indicates the area to the right of Z = 1.40, alternatively, P(Z<-1.40) represents the area to the left of Z = -1.40.
According to the table, P(Z>1.40) = 0.0808, equivalent to 8.08%
In the second question: The probability that a randomly selected education major received a starting salary offer between $45,000 and $52,350 is _______.
Now we must determine the area under the curve connecting the two Z-scores.

We seek the area between Z = -0.30 and Z = 1.40.
This is done via P(Z<1.40) - P(Z < - 0.30) = P(Z > - 1.40) - P(Z < - 0.30)
= 1 - P(Z<-1.40) - P( Z < -0.30)
Now, we need to work with the area under the respective curves relating to Z = - 1.40 and to the left of Z = - 0.30.
From the corresponding table, this evaluation yields: 1 - 0.0808 - 0.3821 = 0.5371
Lastly, for question three: What percentage of education majors received a starting offer between $38,500 and $45,000?
For X = $38,500:

For X = $45,000
Z = -0.3 (calculated above)
Next, we need to evaluate the area under the curve to the right of Z = - 1.80 and to the left of Z = - 0.3
- This results in P (Z > - 1.80) - P (Z < - 0.3)
- 1 - P (Z < - 1.80) - P (Z < - 0.3)
- 1 - 0.0359 - 0.3821 = 0.582 = 58.2%