Respuesta :
Answer:
i) 0.872
ii) 0.300
iii) 0.76
iv) 0.704
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = $261.50 per month
Standard Deviation, σ = $16.25
We are given that the distribution of monthly food cost for a 14- to 18-year-old male is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
a) P(Less than $280)
[tex]P( x < 280) = P( z < \displaystyle\frac{280 - 261.50}{16.25}) = P(z< 1.138)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x < 280) = 0.872 = 87.2\%[/tex]
b) P(More than $270)
P(x > 270)
[tex]P( x > 270) = P( z > \displaystyle\frac{270 - 261.50}{16.25}) = P(z > 0.523)[/tex]
[tex]= 1 - P(z \leq 0.523)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x > 270) = 1 - 0.700 = 0.300 = 30.0\%[/tex]
c) P(More than $250)
P(x > 250)
[tex]P( x > 250) = P( z > \displaystyle\frac{250 - 261.50}{16.25}) = P(z > -0.707)[/tex]
[tex]= 1 - P(z \leq -0.707)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x > 250) = 1 - 0.240 = 0.76 = 76.0\%[/tex]
d) P(Between $240 and $275)
[tex]P(240 \leq x \leq 275) = P(\displaystyle\frac{240 - 261.50}{16.25} \leq z \leq \displaystyle\frac{275-261.50}{16.25}) = P(-1.323 \leq z \leq 0.8307)\\\\= P(z \leq 0.8307) - P(z < -1.323)\\= 0.797 - 0.093 = 0.704 = 70.4\%[/tex]
[tex]P(240 \leq x \leq 275) = 70.4\%[/tex]
e) Thus, 0.704 is the probability that the monthly food cost for a randomly selected 14- to 18-year-old male is between $240 and $275.
