Could someone possibly help me ??
2 answers:
You might be interested in
Answer:
(a) The percentage of the scores were less than 59% is 16%.
(b) The percentage of the scores were over 83% is 2%.
(c) The number of students who received a score over 75% is 26.
Step-by-step explanation:
Let the random variable <em>X</em> represent the scores on a Psychology exam.
The random variable <em>X</em> follows a Normal distribution with mean, <em>μ</em> = 67 and standard deviation, <em>σ</em> = 8.
Assume that the maximum score is 100.
(a)
Compute the probability of the scores that were less than 59% as follows:
*Use a <em>z</em>-table.
Thus, the percentage of the scores were less than 59% is 16%.
(b)
Compute the probability of the scores that were over 83% as follows:
*Use a <em>z</em>-table.
Thus, the percentage of the scores were over 83% is 2%.
(c)
It is provided <em>n</em> = 160 students took the exam.
Compute the probability of the scores that were over 75% as follows:
The percentage of students who received a score over 75% is 16%.
Compute the number of students who received a score over 75% as follows:
Thus, the number of students who received a score over 75% is 26.
Answer: $357 is more amount budgeted for food increase.
Explanation:
Since we have given that
Number of people according to the budget = 150
Cost of food per person= $4.25
Total cost of food is given by
If the number of guests attending the opening was 56% more than expected,
So, number of guests now is given by
Total cost of food is given by
Difference between them is given by
Hence, $357 is more amount budgeted for food increase.