Answer:
0.0323 = 3.23%. Is unusual
Step-by-step explanation:
Using the normal distribution we can find z value with the mean and standard deviation as follows. x is the value we want to know its probability
z = (x - mean)/standard deviation
z = (5-8.54)/1.91
z = -1.85
Using z tables for normal distribution, for z = 1.85, we have an area under the curve of 0.9677. Since we want to know the probability for 5 minutes or less, we have to substract from
p = 1- 0.9677
p = 0.0323
An event with a probability less than 5% is considered unusual.
Steps to solve:
25 = x + 19
~Subtract 19 to both sides
6 = x
Best of Luck!
Answer: 1.25
Step-by-step explanation:
Given: A college-entrance exam is designed so that scores are normally distributed with a mean
= 500 and a standard deviation
= 100.
A z-score measures how many standard deviations a given measurement deviates from the mean.
Let Y be a random variable that denotes the scores in the exam.
Formula for z-score = 
Z-score = 
⇒ Z-score = 
⇒Z-score =1.25
Therefore , the required z-score = 1.25