Respuesta :
Answer:
a) 21.2%
b) $176.05 or more
c) 15.87%
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = $155
Standard Deviation, σ = $25
We are given that the distribution of spending amounts is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
a) P(spends more than $175 per week on groceries)
P(x > 175)
[tex]P( x > 175) = P( z > \displaystyle\frac{175 - 155}{25}) = P(z > 0.8)[/tex]
[tex]= 1 - P(z \leq 0.8)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x > 175) = 1 - 0.788 = 0.212 = 21.2\%[/tex]
b) P(X > x) = 0.2
We have to find the value of x such that the probability is 0.2
P(X > x)
[tex]P( X > x) = P( z > \displaystyle\frac{x - 155}{25})=0.2[/tex]
[tex]= 1 -P( z \leq \displaystyle\frac{x - 155}{25})=0.2 [/tex]
[tex]=P( z \leq \displaystyle\frac{x - 155}{25})=0.8 [/tex]
Calculation the value from standard normal z table, we have,
[tex]P(z < 0.842) = 0.8[/tex]
[tex]\displaystyle\frac{x - 155}{25} = 0.842\\\\x = 176.05[/tex]
A consumer has to spend approximately $176.05 or greater to be in the top 20% of the distribution.
c) My family spends on average $130 dollars on groceries.
P(less than $130)
[tex]P( x < 130) = P( z < \displaystyle\frac{130 - 155}{25}) = P(z < -1)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x < 130) = 0.1587 = 15.87\%[/tex]
Thus, my family percentile is 15.87%
