Question

13. The average price for a home gym setup is $32,560. Assume a normal distribution with σ? = $3870. a) What is the probability that a randomly selected home gym setup price is greater than $41,000?

b) For a sample of 75 home gym setups, what is the probability that the sample price is greater than $38,000?

Answer 1

Solution :

Given that,

(A)P(x > $41000) = 1 - P(x<41000 )

= 1 - P[(x -) / < (41000-32560) / 3870]

= 1 - P(z <2.18 )

Using z table

= 1 -  0.9854

probability=0.0146

(B)

n=75

= / n = 3870 / 75 = 446.8691

P( > 38000) = 1 - P( < 38000)

= 1 - P[( - ) / < (38000-32560) / 446.8691]

= 1 - P(z < 12.17)

Using z table

= 1 -1

= 0

probability=0