Question

The average time spent sleeping (in hours) for a group of healthy adults can be approximated by a normal distribution with a mean of 6.5 hours and a standard deviation of 1 hour. Between what two values does the middle 70% of the sleep times lie?

		5.218 and 7.782
		5.464 and 7.536
		5.658 and 7.342
		4.855 and 8.145

Answer 1

Solution :

Let X be a random variable which represents the average time spent sleeping (in hours) for a group of healthy adults.

Given that,

i.e. Mean (μ) = 6.5 hours

Standard deviation (σ) = 1 hour

Since, normal distribution is symmetrical distribution, therefore middle 70% of the sleep time lie between 15th percentile and 85th percentile.

Let the 15th percentile is k.

Hence, P(X < k) = 0.15

We know that if X ~ N(μ, σ​​​​​​2 ) then

....................(1)

Using "qnorm" function of R we get, P(Z < -1.036) = 0.15

Comparing, P(Z < -1.036) = 0.15 and (1) we get,

The 15th percentile of the sleep time is 5.464 hours.

Let the 85th percentile is c.

Hence, P(X < c) = 0.85

We know that if X ~ N(μ, σ​​​​​​2 ) then

....................(2)

Using "qnorm" function of R we get, P(Z < 1.036) = 0.85

Comparing, P(Z < 1.036) = 0.85 and (2) we get,

The 85th percentile of the sleep time is 7.536 hours.

Hence, middle 70% of the sleep times lie between 5.464 and 7.536 hours.

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