Question

6 THE NORMAL PROBABILITY DISTRIBUTION (26) A large number of random variables observed in nature possess a frequency distribution that is approximately mound-shaped and can be modelled by a normal probability distri- bution (1) True (2) False (27) Suppose is a normal random variable with mean and standard deviation a. If is the standardized normal random variable of I, which of the following statements is false? (1) When the value of 2 = 0. (2) When I is less than the mean y, the value of 2 is negative. (3) When r is greater than the mean, the value of zis positive. (4) It is always the case that <r. Problem F: The waiting time for clients at a dental clinic is normally distributed with an average waiting time of 16.2 minutes and a standard deviation of 3.4 minutes. (28) Refer to Problem F. What is the probability that a client spends less than 14 minutes waiting? (1) 0.1405 (2) 0.5011 (3) 0.2578 (4) 0.1240 (29) Refer to Problem F. What is the probability that a client spends more than 17.5 minutes waiting? (1) 0.6480 (2) 0.3520 (3) 0.5426 (4) 0.2363 (30) Refer to Problem F. What is the probability that a client spends between 15 and 17 minutes waiting? (1) 0.2316 (2) 0.3632 (3) 0.5948 (4) 0.1428

Answer 1

26)

A large number of random variables observed in nature process a frequency distribution that is approximately mound-shaped and it can be modelled by a normal probability distribution

True

27)

The statement that is false is,

It is always the case that z < x

It is not correct because, x can be any random variable that can be in mm values and therefore z is not always less than x.

28)

Given,

u= 16.2

$\small \sigma = 3.4$

We know that,

$\small z = \frac{x-\mu}{\sigma}$

Now,

P(x < 14)

z = (14 - 16.2)/3.4

z = -0.6471

P(z < -0.6471) =  0.2578

29)

P(x > 17.5)

z = (17.5 - 16.2)/3.4

z = 0.3824

P(z < -0.6471) =  0.3520

30)

P(15 < x < 17)

z₁₅ = (15 - 16.2)/3.4 = -0.3529

z₁₇ = (17 - 16.2)/3.4 = 0.2353

P(-0.3529 < x < 0.2353) = 0.2316