Question

Students of a large university spend an average of $6 a day on lunch. The standard deviation of the expenditure is $1.

1. What is the probability that the sample mean will be at least $4.50?
2. What is the probability that the sample mean will be $7.90?
3. Doria spent $2.99 on her lunch on Friday. Explain to her, in terms of standard deviation, why this is not a typical expenditure at this campus.

Answer 1

mean = 6 and sd = 1

using central limit theorem, we find the z-value
z = (xbar - mu)/sigma

a)
P(X >= 4.5)
= P(z > (4.5 - 6)/1)
= P(z > -1.5)
using left tailed z-table
= 0.9332

b)
here we need to find P(X = 7.9)
= P(7.8 < X < 7.9)
= P((7.8 - 4.5)/1 < z < (7.9 - 4.5)/1)
= P(3.3 < z < 3.4)
= P(z < 3.4) - P(z < 3.3)
= 0.0001

c)
lets find the z-score for Doria
z = (2.99 - 6)/1 = -3.01

The value lies more than 3 standard deviation away from the mean to the left of the mean.
Hence this is not a typical expenditure at this campus