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tigry1 [53]
3 years ago
9

The length of time a person takes to decide which shoes to purchase is normally distributed with a mean of 8.54 minutes and a st

andard deviation of 1.91. Find the probability that a randomly selected individual will take less than 5 minutes to select a shoe purchase. Is this outcome unusual?
Mathematics
1 answer:
Minchanka [31]3 years ago
8 0

Answer:

0.0323 = 3.23%. Is unusual

Step-by-step explanation:

Using the normal distribution we can find z value with the mean and standard deviation as follows. x is the value we want to know its probability

z = (x - mean)/standard deviation

z = (5-8.54)/1.91

z = -1.85

Using z tables for normal distribution, for z = 1.85, we have an area under the curve of 0.9677. Since we want to know the probability for 5 minutes or less, we have to substract from

p = 1- 0.9677

p = 0.0323

An event with a probability less than 5% is considered unusual.

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Marta wants to purchase charms for her necklace. Each charm cost $1.59. She wants to spend no more than $20 for the charms. Whic
Svetllana [295]

Which inequality represents this situation ?

1.59×n < 20

How many charms can Marta purchase ?

20 : 1.59 = 12 (+$0.92)

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3 years ago
The test statistic of z= -2.39 is obtained when testing the claim that p&lt;0.52.
krek1111 [17]

Considering the situation described, we have that:

a) The critical value is of z = -1.645.

b) Since the test statistic is less than the critical value, we should reject the null hypothesis H0.

<h3>What is the critical value?</h3>

We have a left-tailed test, as we are testing if the proportion is less than a value. Hence the critical value is z with a p-value equals to the significance level, hence z with a p-value of 0.05, hence z = -1.645.

<h3>What is the decision?</h3>

Considering the test statistic, for a left-tailed test, we have that:

  • Less than the critical value: Reject H0.
  • Equal or greater: Do not reject.

In this problem, z = -2.39 is less than -1.645, hence we should reject the null hypothesis H0.

More can be learned about the test of an hypothesis at brainly.com/question/13873630

#SPJ1

6 0
1 year ago
F(x)=(x+2)^3-4<br> What is the slope?
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Use the slope-intercept y=mx+b to find slope
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4 0
2 years ago
Read 2 more answers
The probability density function of the time to failure of an electronic component in a copier (in hours) is f(x) for Determine
salantis [7]

The question is incomplete. Here is the complete question.

The probability density function of the time to failure of an electronic component in a copier (in hours) is

                                              f(x)=\frac{e^{\frac{-x}{1000} }}{1000}

for x > 0. Determine the probability that

a. A component lasts more than 3000 hours before failure.

b. A componenet fails in the interval from 1000 to 2000 hours.

c. A component fails before 1000 hours.

d. Determine the number of hours at which 10% of all components have failed.

Answer: a. P(x>3000) = 0.5

              b. P(1000<x<2000) = 0.2325

              c. P(x<1000) = 0.6321

              d. 105.4 hours

Step-by-step explanation: <em>Probability Density Function</em> is a function defining the probability of an outcome for a discrete random variable and is mathematically defined as the derivative of the distribution function.

So, probability function is given by:

P(a<x<b) = \int\limits^b_a {P(x)} \, dx

Then, for the electronic component, probability will be:

P(a<x<b) = \int\limits^b_a {\frac{e^{\frac{-x}{1000} }}{1000} } \, dx

P(a<x<b) = \frac{1000}{1000}.e^{\frac{-x}{1000} }

P(a<x<b) = e^{\frac{-b}{1000} }-e^\frac{-a}{1000}

a. For a component to last more than 3000 hours:

P(3000<x<∞) = e^{\frac{-3000}{1000} }-e^\frac{-a}{1000}

Exponential equation to the infinity tends to zero, so:

P(3000<x<∞) = e^{-3}

P(3000<x<∞) = 0.05

There is a probability of 5% of a component to last more than 3000 hours.

b. Probability between 1000 and 2000 hours:

P(1000<x<2000) = e^{\frac{-2000}{1000} }-e^\frac{-1000}{1000}

P(1000<x<2000) = e^{-2}-e^{-1}

P(1000<x<2000) = 0.2325

There is a probability of 23.25% of failure in that interval.

c. Probability of failing between 0 and 1000 hours:

P(0<x<1000) = e^{\frac{-1000}{1000} }-e^\frac{-0}{1000}

P(0<x<1000) = e^{-1}-1

P(0<x<1000) = 0.6321

There is a probability of 63.21% of failing before 1000 hours.

d. P(x) = e^{\frac{-b}{1000} }-e^\frac{-a}{1000}

0.1 = 1-e^\frac{-x}{1000}

-e^{\frac{-x}{1000} }=-0.9

{\frac{-x}{1000} }=ln0.9

-x = -1000.ln(0.9)

x = 105.4

10% of the components will have failed at 105.4 hours.

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3 years ago
Given the speeds of each runner below, determine who runs the fastest.
shutvik [7]

Answer:

Adam is the fastest runner

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3 years ago
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