Question

AS According to a research​ institution, men spent an average of ​$136.78 on​ Valentine's Day gifts in 2009. Assume the standard deviation for this population is ​$45 and that it is normally distributed. A random sample of 15 men who celebrate​ Valentine's Day was selected. Complete A-E

a. Calculate the standard error of the mean.

b. What is the probability that the sample mean will be less than ​$130?

c. What is the probability that the sample mean will be more than $140​?

d. What is the probability that the sample mean will be between ​$115 and ​$165​?

Answer 1

Solution :

Given that,

mean = = $136.78

standard deviation = = $45

a.

n = 15

= $136.78

= / n = 45 / 15 = 11.6190

b.

P( < $130) = P(( - ) / < (130 - 136.78) / 11.6190)

= P(z < -0.5835)

Using standard normal table,

= 0.2798

Probability = 0.2798

c.

P( > $140) = 1 - P( < 140)

= 1 - P(( - ) / < (140 - 136.78) / 11.6190)

= 1 - P(z < 0.2771)

= 1 - 0.6091

= 0.3909

Probability = 0.3909

d.

P($115 < < $165) = P((115 - 136.78) / 11.6190)<( - ) / < (165 - 136.78) / 11.6190))

= P(-1.8745 < Z < 2.4288)

= P(Z < 2.4288) - P(Z < -1.8745) Using z table,

= 0.9924 - 0.0304

= 0.962

Probability = 0.962