Question

i need help with 10 and 12 can i get the steps so i can know for future test.

.8944 10) A sample of size 70 will be drawn from a population with mean 24 and standard deviation 10. Find the probability that will be between 22 and 25. A) 0.7521 B) 0.7326 C) 0.0475 D) 0.2005 10 11) A certain car model has a mean gas mileage of 29 miles per gallon (mpg) with a standard deviation 5 mpg. A pizza delivery company buys 38 of these cars. What is the probability that the average mileage of the fleet is greater than 27.8 mpg? A) 0.0465 B) 0.9306 C) 0.9535 D) 0.8599 12) Use the normal approximation to find the indicated probability. The sample size is n, the population proportion of successes is p, and is the number of successes in the sample. n=96, p=0.42: _/> 45) A) 0.1423 B) 0.8708 C) 0.1292 D) 0.1611 bability. The sample size is n, the

Answer 1

Q10)

Given that ,

$\mu$ = 24

$\sigma$ = 10

A sample of size n = 70 is taken from this population.

Let $url=https://latex.codecogs.com/gif.latex?%5Cbar%20x&t=1575589643&ymreqid=50870fb5-0ab3-84d7-1c55-fb00b101d200&sig=Abcm9Bhpi2BjesKihLBUOQ--~C$ be the mean of sample.

The sampling distribution of the $url=https://latex.codecogs.com/gif.latex?%5Cbar%20x&t=1575589643&ymreqid=50870fb5-0ab3-84d7-1c55-fb00b101d200&sig=Abcm9Bhpi2BjesKihLBUOQ--~C$ is approximately normal with

Mean $\mu_{\bar x}$ = $\mu$ = 24

SD $\sigma_{\bar x}$ =   
ulo
  = 10/$\sqrt{}$​70 =   1.19522860933

Find P(22 < $url=https://latex.codecogs.com/gif.latex?%5Cbar%20x&t=1575589643&ymreqid=50870fb5-0ab3-84d7-1c55-fb00b101d200&sig=Abcm9Bhpi2BjesKihLBUOQ--~C$ < 25)

Z = X - mean /sd

P ( 22-24)/(10/sqrt(70)) < Z < 25-24)/(10/sqrt(70) )

P ( -1.6733 < Z < 0.8366 )

0.7986 - 0.0471

= 0.7515

#Option A is corect

Q11)

mean = $\mu$ = 29

standard deviation = $\sigma$ = 5

n = 38

Find P( $url=https://latex.codecogs.com/gif.latex?%5Cbar%20x&t=1575589643&ymreqid=50870fb5-0ab3-84d7-1c55-fb00b101d200&sig=Abcm9Bhpi2BjesKihLBUOQ--~C$ >27.8)

Z = X - mean /sd

P (z> (27.8-29/(5/sqrt(38))

P (z>-1.4795 )

=0.9306

#OPtion B is corect

Q11)

n=96 p=0.42

mean = $\mu$ =n*p= 40.32

standard deviation = $\sigma$ = sqrt(np(1-p))=4.84

Find P(x>45)

Z = X - mean /sd

P (z> (45-40.32/(4.84)

P (z>0.9677 )

=0.1611

#OPtion D is corect