The time a student sleeps per night has a distribution with mean 6.06 hours and a standard deviation of 0.55 hours. Find the probability that average sleeping time for a randomly selected sample of 35 students is more than 6.15 hours per night.

The probability is 0.16602

Step-by-step explanation:

We start by calculating the z-score

We have this mathematically as;

z-score = (x-mean)/SD/√n

From the question,

we have x = 6.15

mean = 6.06 hours

standard deviation = 0.55 hours

n = 35

Substituting these values, we have

z-score = (6.15-6.06)/0.55/√35

z-score = 0.09/0.55/5.92

z-score = 0.97

So the probability we want to calculate is:

P ( z > 0.97)

we make use of the standard normal distribution table here

The value is; 0.16602

think about it deeply and passionately about this property

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step-by-step explanation: