1. |x+5|=3 How do I get this again?

Assume the heights in a female population are normally distributed with mean 65.7 inches and standard deviation 3.2 inches. Then

maria [59]

Answer:

Choice a. approximately $0.62$ .

Step-by-step explanation:

This explanation shows how to solve this problem using a typical $z$ -score table. Consider a normal distribution with mean $\mu$ and variance $\sigma^2$ . The $z\!$ -score for a measurement of value $x$ would be $(x - \mu) / \sigma$ .

Convert all heights to inches:

$5\; \rm ft = 5 \times 12 \; in = 60\; \rm in$ .
$5\; \rm ft + 7\; \rm in = 5 \times 12 \; in + 7\; \rm in = 67\; \rm in$ .

Let $X$ represent the height (in inches) of a female from this population. By the assumptions in this question: $X \sim \mathrm{N}(65.7,\, 3.2)$ . The question is asking for the probability $P(60 \le X \le 67)$ . Calculate the $z$ score for the two boundary values:

For the lower bound, $60\; \rm in$ : $\displaystyle z = \frac{60 - 65.7}{3.2} \approx -1.78$ .
For the upper bound, $67\; \rm in$ : $\displaystyle z = \frac{67 - 65.7}{3.2} \approx 0.41$ .

Look up the corresponding probabilities on a typical $z$ -score table.

For the $z$ -score of the upper bound, the corresponding probability is approximately $0.6591$ . In other words:

$P(x \le 67) \approx 0.6591$

On the other hand, some $z$ -score table might not include the probability for negative $z\!$ scores. That missing part can be found using the symmetry of the normal distribution PDF.

The probability corresponding to $P(z < 1.78)$ (that's the opposite of the $z\!$ -score at the lower bound) is approximately $0.9625$ . By the symmetry of the normal PDF:

$P(z < -1.78) = 1 - P(z < 0 - (-1.78)) \approx 1 - 0.9625 = 0.0375$ .

Therefore:

$P(X < 60) \approx 0.0375$ .

Calculate the probability of the interval between the two bounds:

$\begin{aligned}P(60 \le X \le 65.7) &= P(X \le 65.7) - P(X \le 60)\\ &\approx 0.6591 - 0.0375 \approx 0.62 \end{aligned}$ .

5 0

4 years ago

The lifespan of a 60-watt lightbulb produced by a company is normally distributed with a mean of 1450 hours and a standard devia

Flura [38]

To evaluate the probability that the lifespan will be between 1440 and 1465 hours will be given by:
P(1440<x<1465)
using the z-score formula we obtain:
z=(x-μ)/σ
where:
μ=1450
σ=8.5
hence
when x=1440
z=(1440-1450)/8.5
z=-1.18
P(z<-1.18)=0.1190

when x=1465
z=(1465-1450)/8.5
z=1.77
P(z<1.77)=0.9625

hence:
P(1440<x<1465)
=0.9625-0.1180
=0.8445

8 0

3 years ago

Q2. Draw the following vectors and determine the resultant

lys-0071 [83]

Spungbib don failed math ph nah