Answer:
Step-by-step explanation:
Hello!
The variable of interest is:
X: number of daily text messages a high school girl sends.
This variable has a population standard deviation of 20 text messages.
A sample of 50 high school girls is taken.
The is no information about the variable distribution, but since the sample is large enough, n ≥ 30, you can apply the Central Limit Theorem and approximate the distribution of the sample mean to normal:
X[bar]≈N(μ;δ²/n)
This way you can use an approximation of the standard normal to calculate the asked probabilities of the sample mean of daily text messages of high school girls:
Z=(X[bar]-μ)/(δ/√n)≈ N(0;1)
a.
P(X[bar]<95) = P(Z<(95-100)/(20/√50))= P(Z<-1.77)= 0.03836
b.
P(95≤X[bar]≤105)= P(X[bar]≤105)-P(X[bar]≤95)
P(Z≤(105-100)/(20/√50))-P(Z≤(95-100)/(20/√50))= P(Z≤1.77)-P(Z≤-1.77)= 0.96164-0.03836= 0.92328
I hope you have a SUPER day!
Answer:
-23/2
Step-by-step explanation:
x-5=3(x+6)
x-3x=18+5
-2x=23
x=-23/2
Answer:
173 MRI machines
Step-by-step explanation:
Margin of error E = 0.5
Confidence interval 90% = 1-0.9 = 0.1
Standard deviation = 4 hours
Number of MRI machines needed per day n, = [(z alpha/2 * SD)/E]²
Z alpha/2 = 1.645 at alpha = 0.1
Inputting these values into n we have that
[(1.645*4)/0.5]²
= 13.16²
= 173.18 is approximately equal to 173
The company has to study 173 machines.