Question 849900: Hi,
I needed help with the following:
The number of minutes per week that students spend surfing the internet is normally distributed with a mean of 490 minutes and standard deviation of 42 minutes.Calculate the probability that the time, in a week, a student spends surfing the internet is
1) more than 500 minutes to 4 dp
2) less than 440 minutes to 4 dp
3) less than 555 minutes to 4 dp
4) between 500 and 555 minutes 0.5342 to 4 dp
5) between 440 minutes and 555 minutes. to 4 dp
6) Given that 67% of students spend less than M minutes,
calculate the value of M, correct to 1 decimal place. to 1 dp
Thank You.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! The number of minutes per week that students spend surfing the internet is normally distributed with a mean of 490 minutes and standard deviation of 42 minutes.Calculate the probability that the time, in a week, a student spends surfing the internet is
1) more than 500 minutes to 4 dp
z(500) = (500-490)/42 = 10/42
P(x > 500) = P(z > 10/42) = normalcdf(10/42,100) = 0.4059
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2) less than 440 minutes to 4 dp
Same procedure using z(440) = (440-490)/42 = -50/42
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3) less than 555 minutes to 4 dp
same
4) between 500 and 555 minutes to 4 dp
z(555)=(555-490)/42 = 65/42
Etc.
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5) between 440 minutes and 555 minutes. to 4 dp
Find P(-50/42< z < 65/42) = normalcdf(-50/42,65/42) = 0.7863
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The number of minutes per week that students spend surfing the internet is normally distributed with a mean of 490 minutes and standard deviation of 42 minutes.Calculate the probability that the time, in a week, a student spends surfing the internet is
1) more than 500 minutes to 4 dp
2) less than 440 minutes to 4 dp
3) less than 555 minutes to 4 dp
4) between 500 and 555 minutes 0.5342 to 4 dp
5) between 440 minutes and 555 minutes. to 4 dp
6) Given that 67% of students spend less than M minutes,
calculate the value of M, correct to 1 decimal place. to 1 dp
Find the z-value with a left-tail of 0.67::
invNorm(0.67) = 0.4399
Find M = 0.4399*42+500 = 518.48
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Cheers,
Stan H.
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