Question

Students of a large university spend an average of $6 a day on lunch. The standard deviation of the expenditure is $2. A simple random sample of 81 students is taken. 1. What is the probability that the sample mean will be at least $5.25? 2. What is the probability that the sample mean will be at least $6.50? 3. What is the range of money spent by people who fall within one standard deviation of the mean? 4. Kelsey spent $12 on her lunch today. Explain to her, in terms of the normal distribution curve and standard deviation, why her purchase is not very typical.

Answer 1

For sampling distribution of mean, P( < A) = P(Z < (A - mean)/standard error)

Mean = $6

Standard error =

=

= 0.2222

1) P(sample mean will be at least $5.25) = P( > 5.25)

= 1 - P( < 5.25)

= 1 - P(Z < (5.25 - 6)/0.2222)

= 1 - P(Z < -3.375)

= 1 - 0.0004

= 0.9996

2) P(sample mean will be at least $6.50) = P( > 6.50)

= 1 - P( < 6.50)

= 1 - P(Z < (6.50 - 6)/0.2222)

= 1 - P(Z < 2.25)

= 1 - 0.9878

= 0.0122

3) Range of money spent within 1 standard deviation of mean = $6 2

= $4 to 8

4) A value is considered usual if it is within 2 standard deviations of mean

(12 - 6)/2 = 3

$12 is 3 standard deviations above mean. So, it is not within the usual range. So, her purchase is not very typical.