1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
bekas [8.4K]
3 years ago
5

The lifetime of a battery in a certain application is normally distributed with mean μ = 16 hours and standard deviation σ = 2 h

ours. What is the probability that a battery will last more than 19 hours?
Mathematics
1 answer:
DaniilM [7]3 years ago
5 0

Answer:

Probability that a battery will last more than 19 hours is 0.0668.

Step-by-step explanation:

We are given that the lifetime of a battery in a certain application is normally distributed with mean μ = 16 hours and standard deviation σ = 2 hours.

<em>Let X = lifetime of a battery in a certain application</em>

So, X ~ N(\mu=16,\sigma^{2} =2^{2})

The z-score probability distribution for normal distribution is given by;

               Z = \frac{  X -\mu}{\sigma}  ~ N(0,1)

where, \mu = mean lifetime = 16 hours

            \sigma = standard deviation = 2 hours

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) 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.

So, the probability that a battery will last more than 19 hours is given by = P(X > 19 hours)

  P(X > 19) = P( \frac{  X -\mu}{\sigma} > \frac{19-16}{2} ) = P(Z > 1.50) = 1 - P(Z \leq 1.50)

                                              = 1 - 0.9332 = 0.0668

<em>Now, in the z table the P(Z </em>\leq<em> x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 1.50 in the z table which has an area of 0.9332.</em>

Hence, the probability that a battery will last more than 19 hours is 0.0668.

You might be interested in
Many, many snails have a one-mile race, and the time it takes for them to finish is approximately normally distributed with mean
Natali [406]

Answer:

The percentage of snails that take more than 60 hours to finish is 4.75%

The relative frequency of snails that take less than 60 hours to finish is .9525

The proportion of snails that take between 60 and 67 hours to finish is 4.52 of 100.

There is a 0% probability that a randomly chosen snail will take more than 76 hours to finish.

To be among the 10% fastest snails, a snail must finish in at most 42.26 hours.

The most typical of 80% of snails that between 42.26 and 57.68 hours to finish.

Step-by-step explanation:

Problems of normally distributed samples can be 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}

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.

In this problem, we have that:

Many, many snails have a one-mile race, and the time it takes for them to finish is approximately normally distributed with mean 50 hours and standard deviation 6 hours.

This means that \mu = 50 and \sigma = 6.

The percentage of snails that take more than 60 hours to finish is %

The pvalue of the zscore of X = 60 is the percentage of snails that take LESS than 60 hours to finish. So the percentage of snails that take more than 60 hours to finish is 100% substracted by this pvalue.

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

Z = \frac{60 - 50}{6}

Z = 1.67

A Zscore of 1.67 has a pvalue of .9525. This means that there is a 95.25% of the snails take less than 60 hours to finish.

The percentage of snails that take more than 60 hours to finish is 100%-95.25% = 4.75%.

The relative frequency of snails that take less than 60 hours to finish is

The relative frequence off snails that take less than 60 hours to finish is the pvalue of the zscore of X = 60.

In the item above, we find that this value is .9525.

So, the relative frequency of snails that take less than 60 hours to finish is .9525

The proportion of snails that take between 60 and 67 hours to finish is:

This is the pvalue of the zscore of X = 67 subtracted by the pvalue of the zscore of X = 60. So

X = 67

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

Z = \frac{67 - 50}{6}

Z = 2.83

A zscore of 2.83 has a pvalue of .9977.

For X = 60, we have found a Zscore o 1.67 with a pvalue of .9977

So, the percentage of snails that take between 60 and 67 hours to finish is:

p = .9977 - 0.9525 = .0452

The proportion of snails that take between 60 and 67 hours to finish is 4.52 of 100.

The probability that a randomly-chosen snail will take more than 76 hours to finish (to four decimal places)

This is 100% subtracted by the pvalue of the Zscore of X = 76.

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

Z = \frac{76 - 50}{6}

Z = 4.33

The pvalue of Z = 4.33 is 1.

So, there is a 0% probability that a randomly chosen snail will take more than 76 hours to finish.

To be among the 10% fastest snails, a snail must finish in at most hours.

The most hours that a snail must finish is the value of X of the Zscore when p = 0.10.

Z = -1.29 has a pvalue of 0.0985, this is the largest pvalue below 0.1. So what is the value of X when Z = -1.29?

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

-1.29 = \frac{X - 50}{6}

X - 50 = -7.74

X = 42.26

To be among the 10% fastest snails, a snail must finish in at most 42.26 hours.

The most typical 80% of snails take between and hours to finish.

This is from a pvalue of .1 to a pvalue of .9.

When the pvalue is .1, X = 42.26.

A zscore of 1.28 is the largest with a pvalue below .9. So

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

1.28 = \frac{X - 50}{6}

X - 50 = 7.68

X = 57.68

The most typical of 80% of snails that between 42.26 and 57.68 hours to finish.

5 0
3 years ago
after eating 25 percent of the pretzels, Sonya had 42 left. how many pretzel did Sonya have originally? ​
Shtirlitz [24]

Answer:

56

Step-by-step explanation:

she ate 25% so she is left with 75%

you want to find that 25% so you do 75÷25=3

42÷3=14

25×4=100

14×4=56

she had 56 pretzels

7 0
3 years ago
I need help with number 1 and 2.
olya-2409 [2.1K]
1.b 2.34 is the correct answer 
8 0
3 years ago
How many rays intersect at point O <br> 1. 5 <br> 2. 4 <br> 3. 6 <br> 4. 3
Vilka [71]

Answer:

3 is correct ans

ray on point o

is 3

option 4 3 is correct

line intersecting on same point on o

5 0
2 years ago
A 50-gallon rain barrel is filled to capacity. It drains at a rate of 10 gallons per minute. Write an equation to show how much
makvit [3.9K]

Answer:

The quantity of water drain after x min is 50 (0.9)^{x}  

Step-by-step explanation:

Given as :

Total capacity of rain barrel = 50 gallon

The rate of drain = 10 gallon per minutes

Let The quantity of water drain after x min = y

Now, according to question

The quantity of water drain after x min = Initial quantity of water ×  (1-\dfrac{\textrm rate}{100})^{\textrm time}

I.e The quantity of water drain after x min = 50 gallon ×  (1-\dfrac{\textrm 10}{100})^{\textrm x}

or,  The quantity of water drain after x min = 50 gallon × (0.9)^{x}

Hence the quantity of water drain after x min is 50 (0.9)^{x}  Answer

4 0
3 years ago
Read 2 more answers
Other questions:
  • 2000lb&gt;1 ton ????????
    13·2 answers
  • Pumps A, B and C operate at their respective constant rates. Pumps A and B, simultaneously, can fill a certain tank in 6/5 hours
    14·1 answer
  • What expression is 15+2(9-4)
    13·1 answer
  • (1) 4p²q : 10pq²
    5·1 answer
  • The picture shows part of a thermometer measuring in degrees Fahrenheit.
    6·1 answer
  • Eddie, 37 year old male, bought a $70,000, 20 year life insurance policy through his employer. Eddie is paid biweekly. How much
    5·2 answers
  • Find x to the nearest tenth.<br><br> A. 0.7<br> B. 1.0<br> C. 1.1<br> D. 1.4
    10·2 answers
  • Let ​f(x)=x2−4 and ​g(x)=7−x. Perform the composition or operation indicated.<br> ​(fg)(8​)
    11·1 answer
  • Company A charges $17, plus $11 per day to rent a piece of equipment. Company B charges $33, plus $9 per day to rent the same pi
    10·1 answer
  • I need answers!!!!!!!!!
    6·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!