Question
I have noticed that students arrive at lecture between 6:55 and 7:10. There are 91 people registered in the class. On average, the percentage of students seated and ready to begin at 7:00 has been 59% (variance = 36) and the distribution has been normally distributed. What is the probability that for a given lecture, there will be fewer than 63% people seated at 7:00?
My Answer:
I drew a normal distribution with the mew being 53.69 (.59*91) and the variance being 6. I calculated the z score and found it was
z= 53.19-53.69/6 = .08= .4681 +.5000= .9681
Im just confused how the probability I came up with would suggest that almost more then 3/4 of the class won't show up at 7 am
In terms of percentages if you mean to say that the distribution is $X \sim Norm(\mu=59, \sigma=6)$ and you are asking for $P(X < 63),$ then the answer is $$P(X < 63) = P((X-\mu)/\sigma) < (63-59)/6) = P(Z < .67) = 0.7486,$$ from normal tables, where some rounding is necessary.
More precisely, without so much rounding, R statistical software (or perhaps a scientific calculator) gives 0.7475.