Respuesta :
Answer:
a) 0.9964
b) 0.3040
Step-by-step explanation:
Given data:
standard deviation = $90,000
Mean sales price =$345,800
sample mean = $370,000
Total number of sample = 100
calculate z score for [/tex](\bar x = 370000)[/tex]
[tex]z = \frac{\bar x - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \frac{370000 - 345800}{\frac{90000}{\sqrt{100}}}[/tex]
z = 2.689
P(x<370000) = P(Z<2.689)
FROM STANDARD NORMAL DISTRIBUTION TABLE FOR Z P(Z<2.689) = 0.9964
B)
calculate z score for (\bar x = 350000)
[tex]z = \frac{\bar x - \mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]z = \frac{350000 - 345800}{\frac{90000}{\sqrt{100}}}[/tex]
z = 2.133
[tex]P(350000 \leq \bar x \leq 365000) = P(0.467 \leq Z\leq 2.133) = P(Z\leq 2.133) - P(Z\leq 0.467)[/tex]
FROM NORMAL DISTRIBUTION TABLE Z VALUE FOR
[tex]P(Z\leq 2.133) = 0.9836[/tex]
[tex]P(Z\leq 0.467) = 0.6796[/tex]
SO, = 0.9836 - 0.6796 = 0.3040
