Answer:
a.) yes
b.) no
c.) yes
d.) yes
Step-by-step explanation:
Answer:
0.0918
Step-by-step explanation:
We know that the average amount of money spent on entertainment is normally distributed with mean=μ=95.25 and standard deviation=σ=27.32.
The mean and standard deviation of average spending of sample size 25 are
μxbar=μ=95.25
σxbar=σ/√n=27.32/√25=27.32/5=5.464.
So, the average spending of a sample of 25 randomly-selected professors is normally distributed with mean=μ=95.25 and standard deviation=σ=27.32.
The z-score associated with average spending $102.5
Z=[Xbar-μxbar]/σxbar
Z=[102.5-95.25]/5.464
Z=7.25/5.464
Z=1.3269=1.33
We have to find P(Xbar>102.5).
P(Xbar>102.5)=P(Z>1.33)
P(Xbar>102.5)=P(0<Z<∞)-P(0<Z<1.33)
P(Xbar>102.5)=0.5-0.4082
P(Xbar>102.5)=0.0918.
Thus, the probability that the average spending of a sample of 25 randomly-selected professors will exceed $102.5 is 0.0918.
ANSWER:
[i] 4 × 10.56 = 42.2
[ii] 5 × 0.432 = 2.16 (not 2.18)
So, <u>Correct choice</u> - [B]
Answer:
gram
Step-by-step explanation:
Answer:
Step-by-step explanation: 7
Subtract 5 on both sides to get 4x = x + 21
then, subtract x from 4x and x is 1, so 3x = 21 then divide by 3 on both sides