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.
36...
<span>divide your 16 by 4, so that you have 1/9. You then multiply the resulting number(4) by your denominator, 9. 4 times 9 = 36</span>
Answer:
27.385 cm longer would we expect Mike's arm span to be than George's.
Step-by-step explanation:
We have to find how many centimeters longer would we expect Mike's arm span to be than George's using the equation:
y= 4.5x + 0.977 x
where y= arm span
and x= height
Given:
Mike's height=x = 175 cm
so Mike's arm span= y= 4.5 x + 0.977 x
= 4.5* (175) + 0.977* (175)
= 787.5 + 170.975
= 958.475 cm
George's height = x = 170 cm
so George's arm span= y= 4.5 x + 0.977 x
= 4.5* (170) + 0.977* (170)
= 765 + 166.09
= 931.09 cm
Mike's arm span longer than George's = Mike's arm span - George's arm span
= 958.475 - 931.09
= 27.385 cm.
