Question

The daily sales of a small retail store in Calgary for the last year are normally distributed with a mean of $2050, and a standard deviation of $300.

What is the probability of daily sales exceeding $2,500?

From a sample of 5 days, what is the probability of having a sample mean less than $2,200?

a. 0.0668; 0.8682

b. 0.4332; 0.0668

c. 0.0668; 0.9332

d. 0.0668; 0.9795

e. 0.4332; Approximately 100%

Answer 1

Solution :

Given that,

mean = = 2050

standard deviation = = 300

P(x >2500 ) = 1 - P(x<2500 )

= 1 - P[(x -) / < (2500-2050) / 300]

= 1 - P(z < 1.5)

Using z table

= 1 - 0.9332

probability= 0.0668

b.

n = 5

= 2050

=  / n = 300 / 5=134.16

P( <2200) = P[( - ) / < (2200-2050) / 134.16]

= P(z < 1.12)

Using z table  

= 0.8682

a. 0.0668; 0.8682