The weights of healthy adult male Labrador Retrievers are approximately Normally distributed with a mean of 77 pounds and a stan
1 answer:
Answer:the photo is for part A.
B) the weight would be Invnorm(.25,77,6)=72.95
C)the weight would be 81.04 but this is the work invnorm(.75,77,6)=81.04
D)IQR=Q3-Q1
81.04-72.95=8.09
E)9.12 is the answer the work normalcdf(85,1000,77,6)=.0912
F).6892 or 68.92% are the answers the work normalcdf(60,80,77,6)=.6892
G).0227 or 2.27% is the answer and the work normalcdf(0,65,77,6)=.0227
Step-by-step explanation:
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Answer:
The probability that a randomly selected depth is between 2.25 m and 5.00 m is 0.55.
Step-by-step explanation:
Let the random variable <em>X</em> denote the water depths.
As the variable water depths is continuous variable, the random variable <em>X</em> follows a continuous Uniform distribution with parameters <em>a</em> = 2.00 m and <em>b</em> = 7.00 m.
The probability density function of <em>X</em> is:
Compute the probability that a randomly selected depth is between 2.25 m and 5.00 m as follows:
Thus, the probability that a randomly selected depth is between 2.25 m and 5.00 m is 0.55.
Answer:
The probability that the last card dealt is an ace is .
Step-by-step explanation:
Given : A deck of ordinary cards is shuffled and 13 cards are dealt.
To find : What is the probability that the last card dealt is an ace?
Solution :
There are total 52 cards.
The total arrangement of cards is 52!.
There is 4 ace cards in total.
Arrangement for containing ace as the 13th card is .
The probability that the last card dealt is an ace is
Therefore, the probability that the last card dealt is an ace is .