A survey of 500 randomly selected customers based on their expenditure patterns was carried out and the data were found to be normally distributed

Question

with a mean of N50 and a standard deviation of N15. Use the information to answer the following questions The probability that any customers are selected at random:

a)         spends more than N80 per week,

           b)    spends less than N50 per week.

The percentage of customers who are expected to:

           c)    spend between N30 and N80 per week,

           d)    spend between N55 and N70 per week.

The expected number of customers who will:

           e)   spend less than N70 per week,

           f)    spend between N37.50 and N57.50 per week

Answer 1

Hi Erick! First off, I'm assuming that the capital Ns of this problem are not intentional and were just a copy-paste issue. If I'm wrong, definitely let me know.

So we've got a normal distribution with mean 50 and standard deviation 15 (that also means variance of 15^2 = 225.)

a) This question requires us to have an understanding of z-scores. Let's calculate the z-score of 80 and then figure out the probability that corresponds to it:

(80-50)/15 = 30/15=2.

Remember, we have to subtract what we find from 1, because we're solving for the probability that someone spends over $80. You can use this command to get the actual probability in R:

1-pnorm(2)

Or, you can look up the number 2 in a z-score table, subtract whatever you find from 1. That should work too. Either way, you should get about 0.023.

b) This one is much easier! If we wanted to, we could go through the same process that we just did. But even better, we can just recognize that 50 is the center of the bell curve. Since the normal distribution is symmetric, half of the curve's area is to the left, and half is to the right. That means the probability is 0.5.

c) For this one, we'll need the z-score corresponding to 80 (already got it above - it's 2) and corresponding to 30: (30-50)/15 = -4/3. We have to solve for this quantity:

P(z<2) - P(z < -4/3),

which we can either use z-tables for or use pnorm(2) - pnorm(-4/3) in R. Either way, you should get about 89%.

d) Same here - we have the Z-score for 55 as (55-50)/15 =1/3 and 70 as (70-50)/15 = 4/3. So we solve

P(z<4/3) - P(z<1/3)

using either a z-table or the pnorm() command in R: pnorm(4/3) - pnorm(1/3). Answer should be about 28%.

e) For this one, all we have to do is find the probability that someone will spend less than 70, and then multiply that probability by the number of people in the sample (in this case 500). Again we get the z-score of 70: (70-50)/15=4/3, and we find P(z<4/3) = .908 by using pnorm(4/3) in R. Multiply that by 500 to get 500*.908 to get about 454 people.

f) Similarly for this one, we'll get the z-scores (57.5-50)/15 = 1/2 and (37.50-50)/15 = -5/6. Solve for the P(z<1/2) - P(z < -5/6) and then multiply by 500. Answer should be about 245 people.