Maths assignment <br> a^3-9a

The set of life spans of an appliance is normally distributed with a mean Mu = 48 months and a standard deviation Sigma = 8 mont

natita [175]

To solve the problem we must know about Z-Score.

<h3>What is Z-score?</h3>

A Z-score helps us to understand how far is the data from the mean. It is a measure of how many times the data is above or below the mean. It is given by the formula,

$Z = \dfrac{X- \mu}{\sigma}$

Where Z is the Z-score,

X is the data point,

μ is the mean and σ is the standard variable.

The life span of an appliance that has a z-score of –3 is 24 months.

Given to us

Mean, μ = 48 months,
standard deviation, σ = 8 months,
Z-Score = -3

<h3> What is the life span of an appliance?</h3>

The life span of the appliance can be calculated using the Z-score formula,

$Z = \dfrac{X- \mu}{\sigma}$

Substitute the values,

$-3 = \dfrac{X- 48}{8}\\\\-3 \times 8 = X - 48\\\\-24+48=X\\\\X = 24\ months$

Hence, the life span of an appliance that has a z-score of –3 is 24 months.

Learn more about Z-score:

brainly.com/question/13299273

6 0

3 years ago

7y-23 plus 23x-16 plus 8x -21 find x and Y show steps please

Solnce55 [7]

7y - 23 + 23x -16 + 8x - 21
31x + 7y - 60
What is the original problem equal to? You can't go any further unless you know.

6 0

3 years ago

17) 100 raffle tickets are sold at $5 a piece.

Softa [21]

The total prize money is 200+100 which is 300

There is 100 tickets sold for $5 each so the people selling the tickets earn $500 and use $300 for prize money.

The expected value of 1 ticket is $5.00

If you still have any questions feel free to comment below

Hope this helped<3

6 0

3 years ago

2. 10 caiete costă 20 lei. Câţi lei costă 7 caiete de acelaşi fel?

lutik1710 [3]

Answer:

Tu vorbesti limba romana! Este foarte cool! Nu am întâlnit niciodată pe cineva care să vorbească română

Step-by-step explanation:

4 0

3 years ago

The lifetime of a certain type of battery is normally distributed with mean value 15 hours and standard deviation 1 hour. There

KIM [24]

Answer:

If the lifetime of batteries in the packet is 40.83 hours or more then, it exceeds for 5% of all packages.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 15

Standard Deviation, σ = 1

Sample size = 4

Total lifetime of 4 batteries = 40 hours

We are given that the distribution of lifetime is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

Standard error due to sampling:

$\displaystyle\frac{\sigma}{\sqrt{n}} = \frac{1}{\sqrt4} = 0.5$

We have to find the value of x such that the probability is 0.05

P(X > x) = 0.05

$P( X > x) = P( z > \displaystyle\frac{x - 40}{0.5})=0.05$

$= 1 -P( z \leq \displaystyle\frac{x - 40}{0.5})=0.05$

$=P( z \leq \displaystyle\frac{x - 40}{0.5})=0.95$

Calculation the value from standard normal z table, we have,

$\displaystyle\frac{x - 40}{0.5} = 1.64\\x = 40.825 \approx 40.83$

Hence, if the lifetime of batteries in the packet is 40.83 hours or more then, it exceeds for 5% of all packages.

8 0

3 years ago

Maths assignment a^3-9a