$\text{The missing part of the question states that we should Note: that N(108,20) model to } \\ \\ \text{ } \text{approximate the distribution of weekly complaints).]}$

Now; assuming X = no of complaints received in a week

Required:

To find P(77 < X < 120)

Using a Gaussian Normal Distribution ( $\mu =$ 108, $\sigma$ = 20)

Using Z scores:

$Z = \dfrac{77-108}{20} \\\\ Z = -\dfrac{35}{20} \\ \\ Z -1.75$

As a result X = 77 for N(108,20) is approximately equal to to Z = -1.75 for N(0,1)

SO;

$Z = \dfrac{120-108}{20} \\ \\ Z = \dfrac{12}{20}\\ \\ Z = 0.6$

Here; X = 77 for a N(108,20) is same to Z = 0.6 for N(0,1)

Now, to determine:

P(-1.75 < Z < 0.6) = P(Z < 0.6) - P( Z < - 1.75)

From the standard normal Z-table:

P(-1.75 < Z < 0.6) = 0.7257 - 0.0401

P(-1.75 < Z < 0.6) = 0.6856

3 0

2 years ago

Which of the following is E ∪ F?

vesna_86 [32]

For E the answer is 2. Because the square of 1 is one and 9 is three. And 2 to the square root of two which would be 2 time 2= 4

5 0

2 years ago

Free Take it Enjoy it​

Free Take it Enjoy it