Question

According to a research​ institution, men spent an average of $132.69 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 10 men who celebrate​ Valentine's Day was selected. Complete parts a through e.

a. Calculate the standard error of the mean.

sigma Subscript x σx=

​(Round to two decimal places as​ needed.)

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

P(x<$130)=

​(Round to four decimal places as​ needed.)

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

P(x>$140)=

​(Round to four decimal places as​ needed.)

d. What is the probability that the sample mean will be between $120 and $165?

(P$120≤x≤$165)=

​(Round to four decimal places as​ needed.)

e. Identify the symmetrical interval that includes​ 95% of the sample means if the true population mean is

​$132.69.

$___≤ x ≤​ $___

(Round to the nearest dollar as​ needed.)

Answer 1

a)

= / sqrt(n) = 45 / sqrt(10) = 14.23

b)

X ~ N ( µ = 132.69 , σ = 45 )
P ( X < 130 )
Standardizing the value
Z = ( X - µ ) / (σ/√(n)
Z = ( 130 - 132.69 ) / ( 45 / √10 )
Z = -0.19
P ( ( X - µ ) / ( σ/√(n)) = ( 130 - 132.69 ) / ( 45 / √(10) )
P ( X < 130 ) = P ( Z < -0.19 )
P ( X̅ < 130 ) = 0.4247

c)

X ~ N ( µ = 132.69 , σ = 45 )
P ( X > 140 ) = 1 - P ( X < 140 )
Standardizing the value
Z = ( X - µ ) / ( σ / √(n))
Z = ( 140 - 132.69 ) / ( 45 / √ ( 10 ) )
Z = 0.51
P ( ( X - µ ) / ( σ / √ (n)) > ( 140 - 132.69 ) / ( 45 / √(10) )
P ( Z > 0.51 )
P ( X̅ > 140 ) = 1 - P ( Z < 0.51 )
P ( X̅ > 140 ) = 1 - 0.695
P ( X̅ > 140 ) = 0.305

d)

X ~ N ( µ = 132.69 , σ = 45 )
P ( 120 < X < 165 )
Standardizing the value
Z = ( X - µ ) / ( σ / √(n))
Z = ( 120 - 132.69 ) / ( 45 / √(10))
Z = -0.89
Z = ( 165 - 132.69 ) / ( 45 / √(10))
Z = 2.27
P ( 120 < X̅ < 165 ) = P ( Z < 2.27 ) - P ( Z < -0.89 )
P ( 120 < X̅ < 165 ) = 0.9884 - 0.1867
P ( 120 < X̅ < 165 ) = 0.8017

e)

X ~ N ( µ = 132.69 , σ = 45 )
P ( a < X < b ) = 0.95
Dividing the area 0.95 in two parts we get 0.95/2 = 0.475
since 0.5 area in normal curve is above and below the mean
Area below the mean is a = 0.5 - 0.475
Area above the mean is b = 0.5 + 0.475
Looking for the probability 0.025 in standard normal table to calculate critical value Z = -1.96
Looking for the probability 0.975 in standard normal table to calculate critical value Z = 1.96
Z = ( X - µ ) / ( σ / √(n) )
-1.96 = ( X - 132.69 ) / ( 45/√(10) )
a = 104.7987
1.96 = ( X - 132.69 ) / ( 45/√(10) )
b = 160.5813

105 < < 161