Respuesta :
Answer:
(A) The probability that a male spent less than $210 online before deciding to visit a store is 0.0668.
(B) The probability that a male spent between $270 and $300 online before deciding to visit a store is 0.0655.
(C) Ninety percent of the amounts spent online by a male before deciding to visit a store is less than $265.632.
Step-by-step explanation:
We are given that the reports indicate that men spend an average of $240 online before they decide to visit a store. If the spending limit is normally distributed and the standard deviation is $20.
Let X = the spending limit
The z-score probability distribution for the normal distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = mean spending limit = $240
[tex]\sigma[/tex] = standard deviation = $20
So, X ~ Normal([tex]\mu=\$240,\sigma^{2} =\$20^{2}[/tex])
(A) The probability that a male spent less than $210 online before deciding to visit a store is given by = P(X < $210)
P(X < $210) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{\$210-\$240}{\$20}[/tex] ) = P(Z < -1.50) = 1 - P(Z [tex]\leq[/tex] 1.50)
= 1 - 0.9332 = 0.0668
The above probability is calculated by looking at the value of x = 1.50 in the z table which has an area of 0.9332.
(B) The probability that a male spent between $270 and $300 online before deciding to visit a store is given by = P($270 < X < $300)
P($270 < X < $300) = P(X < $300) - P(X [tex]\leq[/tex] $270)
P(X < $300) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{\$300-\$240}{\$20}[/tex] ) = P(Z < 3) = 0.9987
P(X [tex]\leq[/tex] $270) = P( [tex]\frac{X-\mu}{\sigma}[/tex] [tex]\leq[/tex] [tex]\frac{\$270-\$240}{\$20}[/tex] ) = P(Z [tex]\leq[/tex] 1.50) = 0.9332
The above probability is calculated by looking at the value of x = 3 and x = 1.50 in the z table which has an area of 0.9987 and 0.9332 respectively.
Therefore, P($270 < X < $300) = 0.9987 - 0.9332 = 0.0655.
(C) Now, we have to find ninety percent of the amounts spent online by a male before deciding to visit a store is less than what value, that is;
P(X < x) = 0.90 {where x is the required value}
P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{x-\$240}{\$20}[/tex] ) = 0.90
P(Z < [tex]\frac{x-\$240}{\$20}[/tex] ) = 0.90
In the z table, the critical value of z that represents the bottom 90% of the area is given as 1.2816, i.e;
[tex]\frac{x-\$240}{\$20}=1.2816[/tex]
[tex]x-240=1.2816\times 20[/tex]
[tex]x=240 + 25.632[/tex]
x = 265.632
Hence, Ninety percent of the amounts spent online by a male before deciding to visit a store is less than $265.632.
