Answer :
Answer: 0.0001
Step-by-step explanation:
Given : For women aged 18-24, systolic blood pressures (in mm Hg) are normally distributed with a mean of 114.8 and a standard deviation of 13.1.
i.e. [tex]\mu=114.8\ \ \ \&\ \ \sigma=13.1[/tex]
Sample size =4
Let x be the sample mean systolic blood pressure.
Then the probability that their mean systolic blood pressure is greater than 140 will be
[tex]P(x>140)=1-P(x\leq140)\\\\=1-P(\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}\leq\dfrac{140-114.8}{\dfrac{13.1}{\sqrt{4}}})\\\\\ =1-P(z\leq3.85)\ \ [\because \ z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}]\\\\=1-0.9999\ \ \text{[By z-table]}\\\\= 0.0001[/tex]
Hence, the required probability = 0.0001
Answer:
Hence, the required probability = 0.0001
Step-by-step explanation: