Answer:
a) 64.26%
b) 4.65%
Step-by-step explanation:
Data Given :
The mean value μ = 26.7
the standard deviation σ = 7.5
a) The probability that that one new compact car selected at random has a fuel economy of at least 24 mpg is determined as follows:
P(x ≥ 24) = 1- P(x < 24)
P(x ≥ 24) = 1- P( X- μ/σ < 24- 26.7/7.5)
P(x ≥ 24) = 1 - P (Z < -0.36)
P(x ≥ 24) = 1 - 0.3594
P(x ≥ 24) = 0.6426
P(x ≥ 24) = 64.26%
b) Assuming If 30 cars were selected
[tex]P(\bar x > 29) = P(\frac{\bar x - \mu }{\frac{\sigma}{\sqrt{n}} } > \frac{29-26.7}{\frac{7.5}{\sqrt{30} } })[/tex]
[tex]P(\bar x > 29) = 1-P(Z < 1.68)[/tex]
By using Z tables ; we have
[tex]P(\bar x > 29) = 1- 0.9535[/tex]
[tex]P(\bar x > 29) =0.0465[/tex]
= 4.65%
