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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 were obtained on a typical day. Researchers found that visually impaired students averaged 9.87 hours of sleep, with a standard deviation of 1.1 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 6.1 hours of sleep?
b. What is the probability that a visually impaired student gets between 6.5 and 7.89 hours of sleep?

Respuesta :

Answer:

a) 0.0003

b) 0.035      

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 9.87 hours

Standard Deviation, σ =  1.1 hours

We are given that the distribution of hours of sleep is a bell shaped distribution that is a normal distribution.

Formula:

[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]

a) P(student gets less than 6.1 hours of sleep)

P(x < 6.1)

[tex]P( x < 6.1) = P( z < \displaystyle\frac{6.1 - 9.87}{1.1}) = P(z < -3.4272)[/tex]

Calculation the value from standard normal z table, we have,  

[tex]P(x < 6.1) = 0.0003[/tex]

c)P(student gets between 6.5 and 7.89 hours of sleep)

[tex]P(6.5 \leq x \leq 7.89) = P(\displaystyle\frac{6.5 - 9.87}{1.1} \leq z \leq \displaystyle\frac{7.89-9.87}{1.1}) = P(-3.063 \leq z \leq -1.8)\\\\= P(z \leq -1.8) - P(z < -3.063)\\= 0.036 - 0.001 = 0.035 = 3.5\%[/tex]

[tex]P(6.1 \leq x \leq 7.89) = 3.5\%[/tex]

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