Principles of statistics

Question

A student has an evening job as a waiter in a restaurant, for which the monthly incomes are normally distributed with a mean of RM 400 and a standard deviation of RM 80.

i. Without doing any calculations, what is the median that the student can earn in a month? Explain your answer.

ii. What is the probability that the student will earn between RM 480 and RM 560 in a
month?

iii. At most, how much will the student earn in 90% of the time?

iv. In a sample of 4 months, on average what is the chance that the student earns between
RM 480 and RM 560 per month?

v. Compare your answers in part (ii) and part (iv). Is there any differences and why?

Answer 1

i) In a normal distribution, the median and mean are equal. In your problem, the mean is 400 so...

 

ii) The normal curve proportions have a general rule 68/95/99.7 rule

    68% of the data will be between 1 standard deviation below to 1 standard deviation above.

    This also means that approximately 34% will be within 1 standard deviation above the mean 

 

   95% of the data will lie between 2 standard deviations below the mean to 2 standard deviations above the mean

 

   That only leaves (95-68)/2 = 27/2 = 13.5% to be between 1 and 2 standard deviations

 

   Here is the breakdown. The top row is the number of standard deviations, the number below represents the        approximate proportion of data falling in that interval

 

<---- -3            -2             -1             0             1             2             3---->

  

  .15%    2.35%       13.5%        34%       34%        13.5%      2.35%     .15%

 

Your question is between 1 and 2 standard deviations since 480 is 400+80   and 560 is 400 + 2(80)

 

Your answer would be approxmately 13.5%

 

 

iii) Using a z-table, 90%tile is associated with 1.28 standard deviations above the mean. Therefore, 400+1.28(80) =502.4  The student will earn RM 502.4 90% of the time

 

iv) and v) The answer would be significantly different because the standard deviation for a sampling distribution of size 4 wouldcut the original standard deviation in half: 80/√4 = 80/2 = 40    σ/√n

 

Now 480 is 2 standard deviations and 560 is 4 standard deviations away from the mean. It is practically equal to the proportion above 2 standard deviations which would be about 2.5%    (I used by calculator and obtained 2.27%)