Question

A​ person's blood pressure is monitored by taking 5 readings daily. The probability distribution of his reading had a mean of 132 and a standard deviation of 3.

a. Each observation behaves as a random sample. Find the mean and the standard deviation of the sampling distribution of the sample mean for the

five observations each day.

b. Suppose that the probability distribution of his blood pressure reading is normal. What is the shape of the sampling​ distribution?

c. Refer to​ (b). Find the probability that the sample mean exceeds 140

Answer 1

Solution :

Given that,

mean = = 132

standard deviation = = 3

n = 5

a) =   = 132

= / n = 3 / 5 = 1.3416

b) The shape of sampling distribution is normal

c) P( > 140) = 1 - P( < 140)

= 1 - P[( - ) / < (140 - 132) / 1.3416 ]

= 1 - P(z < 5.96)

Using z table,    

= 1 - 1

=0

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