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Katena32 [7]
3 years ago
10

Use the following information to determine your answer: The length of a movie falls on a normal distribution. About 95% of movie

s fall between 75 minutes and 163 minutes.
What is the value of the standard deviation for average movie length in minutes? Please round to the second decimal place.
Mathematics
1 answer:
Elenna [48]3 years ago
4 0

Answer:

75= 119 -1.96 \sigma

\sigma = \frac{75-119}{-1.96}= 22.45

And tha's equivalent to use this formula:

163= 119 +1.96 \sigma

\sigma = \frac{163-119}{1.96}= 22.45

Step-by-step explanation:

For this case the 95%of the values are between the following two values:

(75 , 163)

And for this case we know that the variable of interest X "length of a movie" follows a normal distribution:

X \sim N( \mu, \sigma)

We can estimate the true mean with the following formula:

\mu = \frac{75+163}{2}= 119

Now we know that in the normal standard distribution we know that we have 95% of the values between 1.96 deviations from the mean. We can find the value of the deviation with this formula:

75= 119 -1.96 \sigma

\sigma = \frac{75-119}{-1.96}= 22.45

And tha's equivalent to use this formula:

163= 119 +1.96 \sigma

\sigma = \frac{163-119}{1.96}= 22.45

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3 years ago
Suppose an economy experiences an increase in unemployment across all industries. What is the result of this increase in unemplo
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Production would take place at a point inside the production possibility frontier

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Step-by-step explanation:

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Samples of 20 parts from a metal punching process are selected every hour. Typically, 1% of the parts require rework. Let X deno
Roman55 [17]

Answer:

a) P(X>np+3\sqrt{np(1-p)}=0.017

b) P(x>1)=0.190

c) P(Y>1)=0.651

Step-by-step explanation:

This a binomial experiment where success is denoted by parts that need rework.

X ∼ B(n, p); n = 20; p = 0.01

The expected value of X is: E(X) = np =20×0.01= 0.2

The variance is: Var(X) = np(1 − p) = 0.2 × 0.99 = 0.198,

The standard deviation SD(X)= \sqrt{0.198} ≈ 0.445

a) P(X>np+3\sqrt{np(1-p)}=P(X>0.2+3×0.445)=P(X>1.535)=P(X≥2)

Probability function is given by:

\frac{n!}{x!(n-x)!} *p^x*(1-p)^{(n-x)}

P(X≥2)=1-P(X<2)=1-P(X=1)-P(X=0)= 1 - \frac{20!}{1!(20-1)!} *(0.01)^{1}*(1-0.01)^{(20-1)}-\frac{20!}{0!(20-0)!} *(0.01)^{0}*(1-0.01)^{(20-0)}

P(X≥2)=1-0.165-0.818=0.017

b) p=0.04

P(x>1)=P(x≥2)= 1 - P(x=1) - P(x=0)= 1 - \frac{20!}{0!(20-1)!} *(0.04)^{1}*(1-0.04)^{(20-1)} - \frac{20!}{0!(20-0)!} *(0.04)^{0}*(1-0.04)^{(20-0)}

P(x>1)= 1 - 0.368 - 0.442=0.190

c) In this case we consider p=0.19 (Probability that X exceeds 1)

In this experiment Y is the number of hours and n= 5 hours.

Then, we check the probability in each hour:

P(Y>1)=1- P(Y=0)

P(Y=0)=\frac{5!}{0!(5-0)!} *(0.19)^{0}*(1-0.19)^{(5-0)}=0.349

P(Y>1)=1-0.349=0.651

3 0
3 years ago
Whats is 4 hundreds 13 tens 5 ones in standard form
mars1129 [50]
<span>4 hundreds 13 tens 5 ones = 400+13*10+5
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