Ask question

Ask question

All categories

2 years ago

9

A GE lightbulb is supposed to last for 1,200 hours. In fact, light bulbs of this type last only 1,185 hours with a standard devi

ation of 70 hours. What is the probability that a sample of 100 lightbulbs will have an average life of at least 1,200 hours? Mathematics

1 answer:

dybincka [34]2 years ago

7 0

This can be solved using the formula for t test which is = x - mu over standard deviation divided by the square root of the number of samples. In this case, it is 1185 - 1200 over 70 divided by square root of 100 which give us a value of 2.14 which is equal to .9839 in area. Only rarely, just over one time in a hundred tries of 100 light bulbs, would the average life exceed 1200 hours.

You might be interested in

Luke has 1/5 of a package of dried apricots. He divides the dried apricots equally into 4 small bags. Luke gives one of the bags

Ivahew [28]

Answer:

3/80

Step-by-step explanation:

If one fifth of apricots are split into 4 parts, each bag has 1/5 * 1/4 of the original apricots

1/5 * 1/4 = 1/20

Luke keeps 3/4 of those so that's

3/4 * 1/20 = 3/80

7 0

1 year ago

University graduates have a mean job search time of 38.1 weeks, with a standard deviation of 10.1 weeks. The distribution of job

Lera25 [3.4K]

Answer:

7.1 weeks to 68.4 weeks

Step-by-step explanation:

Chebyshev's Theorem states that:

75% of the measures are within 2 standard deviations of the mean.

89% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean = 38.1

Standard deviation = 10.1

Between what two search times does Chebyshev's Theorem guarantee that we will find at least 89% of the graduates

Between 3 standard deviations of the mean.

So from 38.1 - 3*10.1 = 7.8 weeks to 38.1 + 3*10.1 = 68.4 weeks

5 0

1 year ago

The Census Bureau reports that 82% of Americans over the age of 25 are high school graduates. A survey of randomly selected resi

SVETLANKA909090 [29]

Answer:

a) Mean = 1030; Standard deviation = 12.38.

b) The county result is unusually high.

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. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

(a) Find the mean and standard deviation for the number of high school graduates in groups of 1210 Americans over the age of 25.

This first question is a binomial propability distribution.

We have a sample of 1210 Amricans, so $n = 1210$ .

The mean of the sample is 1030.

The probability of a success is $\pi = \frac{1030}{1210} = 0.8512$ .

The standard deviation of the sample is $s = \sqrt{n\pi(1-\pi)} = \sqrt{1210*0.8512*0.1488} = 12.38$

(b) Is that county result of 1030 unusually high, or low, or neither?

The first step is find the zscore when $X = 1030$ .

Then we find the pvalue of this zscore.

If this pvalue is bigger than 0.95, the county result is unusually high.

If this pvalue is smaller than 0.05, the county result is unusually low.

Otherwise, it is neither.

The national mean is 82%. So,

$\mu = 0.82(1210) = 992.2$

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

$Z = \frac{1030 - 992.2}{12.38}$

$Z = 3.05$

$Z = 3.05$ has a pvalue of 0.9989.This means that the county result is unusually high.

4 0

2 years ago

A random sample of 28 statistics tutorials was selected from the past 5 years and the percentage of students absent from each on

SVEN [57.7K]

Answer:

Step-by-step explanation:

Hello!

X: number of absences per tutorial per student over the past 5 years(percentage)

X≈N(μ;σ²)

You have to construct a 90% to estimate the population mean of the percentage of absences per tutorial of the students over the past 5 years.

The formula for the CI is:

X[bar] ± $Z_{1-\alpha /2}$ * $\frac{S}{\sqrt{n} }$

⇒ The population standard deviation is unknown and since the distribution is approximate, I'll use the estimation of the standard deviation in place of the population parameter.

Number of Absences 13.9 16.4 12.3 13.2 8.4 4.4 10.3 8.8 4.8 10.9 15.9 9.7 4.5 11.5 5.7 10.8 9.7 8.2 10.3 12.2 10.6 16.2 15.2 1.7 11.7 11.9 10.0 12.4

X[bar]= 10.41

S= 3.71

$Z_{1-\alpha /2}= Z_{0.95}= 1.645$

[10.41±1.645* $(\frac{3.71}{\sqrt{28} } )$ ]

[9.26; 11.56]

Using a confidence level of 90% you'd expect that the interval [9.26; 11.56]% contains the value of the population mean of the percentage of absences per tutorial of the students over the past 5 years.

I hope this helps!

7 0

2 years ago

The number of airline passengers in 1990 was 466 million. The number of passengers traveling by airplane each year has increased

taurus [48]

Answer:

2010.

Step-by-step explanation:

We have been given an exponential growth formula $P(t)=466\cdot 1.035^t$ , which represents number of passengers traveling by airplane since 1990.

To find the year in which 900 million passengers will travel by airline, we will equate the given formula by 900 and solve for t as:

$900=466\cdot 1.035^t$

$\frac{900}{466}=\frac{466\cdot 1.035^t}{466}$ $1.9313304721030043=1.035^t$

Take natural log of both sides:

$\text{ln}(1.9313304721030043)=\text{ln}(1.035^t)$

Using property $\text{ln}(a^b)=b\cdot \text{ln}(a)$ , we will get:

$\text{ln}(1.9313304721030043)=t\cdot \text{ln}(1.035)$

$0.658209129198=t\cdot 0.034401426717$

$\frac{0.658209129198}{0.034401426717}=\frac{t\cdot 0.034401426717}{0.034401426717}$

$19.1331927775=t\\\\t=19.1331927775$

This means that in the 20th year since 1990, 900 million passengers would travel by airline.

$1990+20=2010$

Therefore, 900 million passengers would travel by airline in 2010.

6 0

1 year ago