Mathematics, 04.01.2020 05:31 coltonwsmith

The weight of people on a college campus are normally distributed with mean 185 pounds and standard deviation 20 pounds. what's the probability that a person weighs more than 200 pounds? (round your answer to the nearest hundredth)

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Answer from: elenadelfosse

0.23.

Step-by-step explanation:

We have been given that the weight of people on a college campus are normally distributed with mean 185 pounds and standard deviation 20 pounds.

First of all, we will find the z-score corresponding to sample score 200 using z-score formula.

$z=\frac{x-\mu}{\sigma}$ , where,

Z-score,

Sample score,

$\mu=$ Mean,

$\sigma=$ Standard deviation.

$z=\frac{200-185}{20}$

$z=\frac{15}{20}$

z=0.75

Now, we need to find P(z0.75) . Using formula P(za)=1-P(z , we will get:

P(z0.75)=1-P(z

Using normal distribution table, we will get:

P(z0.75)=1-0.77337

P(z0.75)=0.22663

Round to nearest hundredth:

$P(z0.75)\approx 0.23$

Therefore, the probability that a person weighs more than 200 pounds is approximately 0.23.

Answer from: emmaguentherp3hjd3

the probability that a person weighs more than 200 pounds is 0.23

Step-by-step explanation:

Since the weight of people on a college campus are normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - u)/s

Where

x = weight of people on a college campus

u = mean weight

s = standard deviation

From the information given,

u = 185

s = 20

We want to find the probability that a person weighs more than 200 pounds. It is expressed as

P(x greater than 200) = P(x greater than 200) = 1 - P(x lesser than lesser than or equal to 200).

For x = 200,

z = (200 - 185)/20 = 0.75

Looking at the normal distribution table, the probability corresponding to the z score is 0.7735

P(x greater than 200) = 1 - 0.7735 = 0.23

Answer from: Quest

geometric series converge the common ratio r has magnitude less than one.

$|r| < 1$

which is equivalent to

$-1 < r < 1$

for the example we need

|2-x| < 1

-1 < 2-x and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1 diverges

1b. r=50/100=1/2 converges

when these converge they converge to

$s = \dfrac{a}{1 -r}$

where a is the first term.

$s = \dfrac{100}{1 - 1/2} = 200$

answer: 200

1c. r=-1 so it's not the case that |r| < 1. diverges

2a. x is the common ratio so we converge when |x|< 1

2b. r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c. r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}

$|x| < \frac 1 2 \sqrt[3]{2}$

Answer from: Quest

you never say the total so we cant find the percentage used for petrol

step-by-step explanation:

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