Question

(a) draw and label a sketch of the normal curve

(b) identify and shade the area of interest

(c) identify any formulas and values substituted

(d) identify the calculator command used and values entered into the calculator

(e) write your response as a decimal rounded to three places

A greenhouse in a tri-county area has kept track of its customers for the last several years and has determined that 28% of its customers plant a vegetable garden in the spring.  

Use the Central Limit Theorem.

a.  In a random sample of 1000 customers, what is the probability that at most 300 customers plant a vegetable garden in the spring.

b.   In a random sample of 1000 customers, what is the probability that at least 225 customers plant a vegetable garden in the spring?

c.   In a random sample of 1000 customers, what is the probability that 275 to 325 customers plant a vegetable garden in the spring?

Answer 1

p = 0.28

According to central limit theorem, sampling distribution of sample proportion is normally distributed with mean = and standard deviation is

if n*p > 10, n*(1-p) > 10

n = 1000

p = 0.28

n * p = 1000 * 0.28 = 280

n * (1 - p) = 1000 * (1-0.28) = 1000 * 0.72 = 720 > 10

Conditions are satisfied.

So we can say that the sampling distribution of sample proportion is normally distributed with mean = and standard deviation is

(Round to 4 decimal)

a)

Here we have to find the probability that at most 300 customers plant a vegetable garden in the spring that is at most 300/1000 = 0.3 proportion of customers plant a vegetable garden in the spring

   where z is standard normal variable.

   (Round to 2 decimal)

= 0.9207 (From statistical table of z values)

The probability that at most 300 customers plant a vegetable garden in the spring is 0.9207

b)

Here we have to find probability that at least 225 customers plant a vegetable garden in the spring that is at least 225/1000 = 0.225 proportion of customers plant a vegetable garden in the spring

   where z is standard normal variable.

(Round to 2 decimal)

= 1 - P(z < -3.87)

= 1 - 0 (From statistical table of z values)

= 1

The probability that at least 225 customers plant a vegetable garden in the spring is 1

c)

Here we have to find the probability that 275 to 325 customers plant a vegetable garden in the spring that is 275/1000 = 0.275 to 325/1000 = 0.325 proportion of customers plant a vegetable garden in the spring

     

  

(Round to 2 decimal)

= P(z < 3.17) - P(z < -0.35)

= 0.9992 - 0.3632    (From statistical table of z values)

= 0.6361

The probability that 275 to 325 customers plant a vegetable garden in the spring is 0.6361