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Alisiya [41]
3 years ago
7

The average amount of time that students use computers at a university computer center is 36 minutes with a standard deviation o

f 5 minutes. The times are known to be normally distributed. Around 10,000 uses are recorded each week in the computer center. The computer center administrative committee has decided that if more than 2000 uses of longer than 40 minutes at each sitting are recorded weekly, some new terminals must be purchased to meet usage needs. Should the computer center purchase the new computers? (Round your percent to two decimal places, and round the number of computer users to the nearest user.)
Mathematics
1 answer:
frosja888 [35]3 years ago
8 0

Answer:

2119 students use the computer for more than 40 minutes. This number is higher than the threshold estabilished of 2000, so yes, the computer center should purchase the new computers.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 36, \sigma = 5

The first step to solve this question is finding the proportion of students which use the computer more than 40 minutes, which is 1 subtracted by the pvalue of Z when X = 40. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{40 - 36}{5}

Z = 0.8

Z = 0.8 has a pvalue of 0.7881.

1 - 0.7881 = 0.2119

So 21.19% of the students use the computer for longer than 40 minutes.

Out of 10000

0.2119*10000 = 2119

2119 students use the computer for more than 40 minutes. This number is higher than the threshold estabilished of 2000, so yes, the computer center should purchase the new computers.

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The salary of teachers in a particular school district is normally distributed with a mean of $50,000 and a standard deviation o
andrey2020 [161]

Answer:

D. 54,900

Step-by-step explanation:

We have been given that the salary of teachers in a particular school district is normally distributed with a mean of $50,000 and a standard deviation of $2,500.

To solve our given problem, we need to find the sample score using z-score formula and normal distribution table.

First of all, we will find z-score corresponding to probability 0.975(1-0.025) using normal distribution table.  

From normal distribution table, we get z-score corresponding is 1.96.

Now, we will use z-score formula to find sample score as:

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

z = Z-score,

z = Sample score,

\mu = Mean,

\sigma = Standard deviation

1.96=\frac{x-50,000}{2,500}

1.96*2,500=\frac{x-50,000}{2,500}*2,500

4900=x-50,000

4900+50,000=x-50,000+50,000

54900=x

Therefore, the salary of $54900 divides the teachers into one group that gets a raise and one that doesn't.

4 0
3 years ago
Godfrey plays a game in which he throws two fair six sided dice. If he rolls two sizes, he wins 20p, if he rolls one six, he win
blagie [28]

Answer:

The Probabilty distribution for the amount Godfrey gains in one turn is then given as

X ||| P(X)

15p | 0.0278

5p | 0.278

-5p | 0.6942

Step-by-step explanation:

If random variable X represents the amount Godfrey gains in one turn.

There are 3 different possible outcomes for X.

- Godfrey pays 5p to enter the game and gets two sixes and wins 20p.

Net gain = 15p

Probability of getting two sixes from two fair dice

= (number of outcomes with two sixes) ÷ (total number of outcomes)

number of outcomes with two sixes = 1

total number of possible outcomes = 36

Probability of getting two sides from two fair dice = (1/36) = 0.0278

- Godfrey pays 5p to enter the game and gets only one six and wins 10p.

Net gain = 5p

Probability of getting one six from either of two fair dice

= (number of outcomes with one six) ÷ (total number of outcomes)

number of outcomes with one six = 2 × n[(6,1), (6,2), (6,3), (6,4), (6,5)] = 2 × 5 = 10

total number of possible outcomes = 36

Probability of getting two sides from two fair dice = (10/36) = 0.278

- Godfrey pays 5p to enter the game and doesn't win anything

Net gain = -5p

Probability of not getting two sixes or one six.

= 1 - [(Probability of getting two sixes) + (Probability of getting one six on.wither dice)]

= 1 - 0.0278 - 0.278 = 0.6942

Probability of getting not getting two sixes or one six = 0.6942

The Probabilty distribution for the amount Godfrey gains in one turn is then given as

X ||| P(X)

15p | 0.0278

5p | 0.278

-5p | 0.6942

Hope this Helps!!!

4 0
4 years ago
Read 2 more answers
Given the function f(x) =2x+5 what would the output be with an input of 6?
Travka [436]
The output would be 17
6 0
3 years ago
A regular hexagon (6-sided figure) has sides that are all the same length. Each side length is represented by the expression x +
Mandarinka [93]
Perimiter=6 times legnth of one side
P=6(x+1/3y)
P=6x+6/3y
P=6x+2y

if x=5 and y=6
P=6(5)+2(6)
P=30+12
P=42


A. P=6x+2y
B. 42 units is the perimiter
8 0
4 years ago
10 POINTS PLEASE HELP!!! IT SEEMS EASY BUT I DONT UNDERSTAND!!!!
Leni [432]
Hello there!

I’m not sure if they want me to add the answers together or just do them separately.

Exponents
(-2)^2=(-2)(-2)=4
(7/6)^2=7/6(7/6)=49/36

I hope this helps!
Best wishes.
-HuronGirl

3 0
4 years ago
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