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3 years ago

The heights of women in the USA are normally distributed with a mean of 64 inches and a standard deviation of 3 inches.

Mathematics

1 answer:

Rainbow [258]3 years ago

8 0

Answer:

(a) 0.2061

(b) 0.2514

Step-by-step explanation:

Let X denote the heights of women in the USA.

It is provided that X follows a normal distribution with a mean of 64 inches and a standard deviation of 3 inches.

(a)

Compute the probability that the sample mean is greater than 63 inches as follows:

$P(\bar X>63)=P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}>\frac{63-64}{3/\sqrt{6}})\\\\=P(Z>-0.82)\\\\=P(Z$

Thus, the probability that the sample mean is greater than 63 inches is 0.2061.

(b)

Compute the probability that a randomly selected woman is taller than 66 inches as follows:

$P(X>66)=P(\frac{X-\mu}{\sigma}>\frac{66-64}{3})\\\\=P(Z>0.67)\\\\=1-P(Z$

Thus, the probability that a randomly selected woman is taller than 66 inches is 0.2514.

(c)

Compute the probability that the mean height of a random sample of 100 women is greater than 66 inches as follows:

$P(\bar X>66)=P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}>\frac{66-64}{3/\sqrt{100}})\\\\=P(Z>6.67)\\\\\ =0$

Thus, the probability that the mean height of a random sample of 100 women is greater than 66 inches is 0.

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Outside temperature over a day can be modeled as a sinusoidal function. Suppose you know the temperature varies between 73 and 9

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Answer: After about 9.03 hours the temperature first reach 82 degrees.

Step-by-step explanation:

The sinusoidal function is given by :

$y=A\sin[\omega(x-\alpha)]+C$

where, A = amplitude; $\omega=\dfrac{2\pi}{period}$ , α= phase shift on the Y-axis and C = midline.

As per given,

Average daily temperature= $C=\dfrac{73+97}{2}=85$ [midline is average of upper and lower limit.]

A= 97-85 = 12

Phase shift: $\alpha=10$

Period = 24 hours;

$\omega=\dfrac{2\pi}{24}=\dfrac{\pi}{12}$

Substitute all values in sinusoidal function, we get

$y=12\sin[\dfrac{\pi}{12}(x-10)]+85$

Put y= 82, we get

$82=12\sin[\dfrac{\pi}{12}(x-10)]+85\\\\\Rightarrow\ -3= 12\sin[\dfrac{\pi}{12}(x-10)]\\\\=\dfrac{-1}{4}= \sin[\dfrac{\pi}{12}(x-10)]\\\\\Rightarrow\ \dfrac{\pi}{12}(x-10)=\sin^{-1}(\dfrac{-1}{4})\\\\\Rightarrow\ x-10=\dfrac{12}{\pi}(\sin^{-1}(\dfrac{-1}{4}))\\\\\Rightarrow\ x=\dfrac{12}{\pi}(\sin^{-1}(\dfrac{-1}{4}))+10\\\Rightarrow\ x\approx9.03$