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Cerrena [4.2K]
3 years ago
6

​john's commute time to work during the week follows the normal probability distribution with a mean time of 26.7 minutes and a

standard deviation of 5.1 minutes. what is the probability that the commute time for a randomly selected day will be between 28 and 34​ minutes?
Mathematics
1 answer:
Angelina_Jolie [31]3 years ago
6 0
To evaluate the probability that a randomly selected day will be between 28 and 34 minutes we proceed as follows:
P(28<x<34)
First we evaluate the z-score for the above values:
z=(x-σ)/μ
μ=26.7
σ=5.1

when:
x=28
z=(28-26.7)/5.1
z=0.26
P(z<0.26)=0.6026

when x=34
z=(34-26.7)/5.1
z=1.43
P(z<1.43)=0.9236
hence:
P(28<x<34)=0.9236-0.6026=0.321~32.1%

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Answer:

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Step-by-step explanation:

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Katie practice the flute for 45 minutes then she ate a snack for 15 minutes next she watched television for for 30 minutes until
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Last week it snowed 4.32 inches on Monday and 6.86 inches on Tuesday. It snowed 7.89 inches this week.
a_sh-v [17]

Answer:

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4 years ago
A researcher wishes to estimate the percentage of adults who support abolishing the penny. What size sample should be obtained i
alex41 [277]

Answer:

sample size if he uses a previous estimate of 32​% is 523

sample size if he does not use any prior​ estimates is 601

Step-by-step explanation:

estimate E = 4 percentage = 0.04

confidence Cl = 95% = 0.95

previous estimate p = 32% = 0.32

q = 1 - 0.32 = 0.68

to find out

size of sample

solution

first we calculate z value for 95% confidence E = 0.04 is 1.96

from probability P(-1.96 < z < 1.96) = 0.95)

here z = 1.96

so in 1st part size of sample we know

E = z × \sqrt{pq/n}

put all value E, z p and q and we get n

n = (z/E)² ×p×q

n = (1.96/0.04)² ×0.32×0.68

n = 523

sample size if he uses a previous estimate of 32​% is 523

now in 2nd part we take p = 0.5

so n will be

n = (z/E)² ×p×q

n = (1.96/0.04)² ×0.5×0.5

n = 601

sample size if he does not use any prior​ estimates is 601

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