At a picnic, a 16 feet long sandwich is cut into 2 pieces. The ratio of the lengths of the 2 pieces is 3:1. What is the length,t
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Answer:
(a) The standard deviation of the amount spent is $3229.18.
(b) The probability that a household spends between $4000 and $6000 is 0.2283.
(c) The range of spending for 3% of households with the highest daily transportation cost is $12382.86 or more.
Step-by-step explanation:
We are given that the average annual amount American households spend on daily transportation is $6312 (Money, August 2001). Assume that the amount spent is normally distributed.
(a) It is stated that 5% of American households spend less than $1000 for daily transportation.
Let X = <u><em>the amount spent on daily transportation</em></u>
The z-score probability distribution for the normal distribution is given by;
Z = ~ N(0,1)
where, = average annual amount American households spend on daily transportation = $6,312
= standard deviation
Now, 5% of American households spend less than $1000 on daily transportation means that;
P(X < $1,000) = 0.05
P( <
) = 0.05
P(Z < ) = 0.05
In the z-table, the critical value of z which represents the area of below 5% is given as -1.645, this means;
= 3229.18
So, the standard deviation of the amount spent is $3229.18.
(b) The probability that a household spends between $4000 and $6000 is given by = P($4000 < X < $6000)
P($4000 < X < $6000) = P(X < $6000) - P(X $4000)
P(X < $6000) = P( <
) = P(Z < -0.09) = 1 - P(Z
0.09)
= 1 - 0.5359 = 0.4641
P(X $4000) = P(
) = P(Z
-0.72) = 1 - P(Z < 0.72)
= 1 - 0.7642 = 0.2358
Therefore, P($4000 < X < $6000) = 0.4641 - 0.2358 = 0.2283.
(c) The range of spending for 3% of households with the highest daily transportation cost is given by;
P(X > x) = 0.03 {where x is the required range}
P( >
) = 0.03
P(Z > ) = 0.03
In the z-table, the critical value of z which represents the area of top 3% is given as 1.88, this means;
x = $6312 + 6070.86 = $12382.86
So, the range of spending for 3% of households with the highest daily transportation cost is $12382.86 or more.
Check the tree diagram, of all the possible scenarios and the probabilities.
All the possible scenarios are:
1st.non defective, 2nd.non defective
1st.non defective, 2nd. defective
1st. defective, 2nd. non defective
1st. defective, 2nd. defective
consider the case 1st.non defective, 2nd.non defective
the probability that the first one is non defective is 8/10, since 8 out of 10 are non defective.
In this scenario, the second one is non defective as well. The probability for this to happen is 7/9, since now we have 7 non defective pots, and 9 in total.
This means that the probability of the first to be non defective, and the second non-defective is (8/10) * (7/9)=0.622
similarly we calculate the other cases, as shown in the picture.
P(at least one is defective)=
P(1st.non defective, 2nd. defective)+P(1st. defective, 2nd. non defective)+P(1st. defective, 2nd. defective)=
0.178+0.178+0.022=0.378
Answer: 0.378
All the possible scenarios are:
1st.non defective, 2nd.non defective
1st.non defective, 2nd. defective
1st. defective, 2nd. non defective
1st. defective, 2nd. defective
consider the case 1st.non defective, 2nd.non defective
the probability that the first one is non defective is 8/10, since 8 out of 10 are non defective.
In this scenario, the second one is non defective as well. The probability for this to happen is 7/9, since now we have 7 non defective pots, and 9 in total.
This means that the probability of the first to be non defective, and the second non-defective is (8/10) * (7/9)=0.622
similarly we calculate the other cases, as shown in the picture.
P(at least one is defective)=
P(1st.non defective, 2nd. defective)+P(1st. defective, 2nd. non defective)+P(1st. defective, 2nd. defective)=
0.178+0.178+0.022=0.378
Answer: 0.378