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Naddika [18.5K]
3 years ago
14

Using diaries for many weeks, a study on the lifestyles of visually impaired students was conducted. The students kept track of

many lifestyle variables including how many hours of sleep obtained on a typical day. Researchers found that visually impaired students averaged 8.65 hours of sleep, with a standard deviation of 1.93 hours. Assume that the number of hours of sleep for these visually impaired students is normally distributed.
(a) What is the probability that a visually impaired student gets less than 6.2 hours of sleep?

answer:

(b) What is the probability that a visually impaired student gets between 6.8 and 10.93 hours of sleep?

answer:

(c) Thirty percent of students get less than how many hours of sleep on a typical day?

answer: hours
Mathematics
1 answer:
Hoochie [10]3 years ago
3 0

Answer:

a) 0.1020

b) 0.7125

c) 7.64 hours

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}

The Z-score measures how many standard deviations the measure is from the mean. 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, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 8.65, \sigma = 1.93

(a) What is the probability that a visually impaired student gets less than 6.2 hours of sleep?

This is the pvalue of Z when X = 6.2. So

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

Z = \frac{6.2 - 8.65}{1.93}

Z = -1.27

Z = -1.27 has a pvalue of 0.1020.

So the answer for a is 0.1020.

(b) What is the probability that a visually impaired student gets between 6.8 and 10.93 hours of sleep?

This is the pvalue of Z when X = 10.93 subtracted by the pvalue of Z when X = 6.8. So

X = 10.93

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

Z = \frac{10.93 - 8.65}{1.93}

Z = 1.18

Z = 1.18 has a pvalue of 0.8810.

X = 6.8

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

Z = \frac{6.8 - 8.65}{1.93}

Z = -0.96

Z = -0.96 has a pvalue of 0.1685.

So the answer for b) is 0.8810 - 0.1685 = 0.7125.

(c) Thirty percent of students get less than how many hours of sleep on a typical day?

This is the value of X in the 30th percentile, that is, the value of X when Z has a pvalue of 0.30. So it is X when Z = -0.525, since this happens between Z = -0.53 and Z = -0.52.

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

-0.525 = \frac{X - 8.65}{1.93}

X - 8.65 = -0.525*1.93

X = 7.64

The answer for c is 7.64 hours.

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Greatest common factor (or denominator) is 6,
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hmm
(note: I spent like 30 mins trying to use a math only of finding the values but it didn't work so I did a  force brute and elimination method explained below)


so
a and b must be multiples of 6
so list all the multiples of 6
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180=6*30 and 30's factors are 1,2,3,5,6,10,15,30 so only list the numbers that are the result of multiplying 6 and any of those numbers in that list (so we can have the lcm of 180)
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6*1=6
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now we must find the pair that has a GCD of only 6

doing the math is long and tedious so do it yourself (trial and error)

we see that our choices that fulfill both requirements (GCD of 6 and LCM of 180) are
90&12
60&18
30&36
sum them to find the least one

90+12=102
60+18=78
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