Daily water intake (including water used in drinks such as coffee, tea and juice) for Canadian adults follows a normal distribution with mean 1.86 litres and standard deviation 0.29 litres.
(a) Can you calculate the probability that the mean daily water intake for a random sample of two Canadian adults is less than 1.65 litres? If you can, enter the probability in the answer box. If you can't, type the word No.
(b) What is the probability that the mean daily water intake for a random sample of 31 Canadian adults is greater than 1.8 litres?
(c) What is the probability that the total daily water intake for a random sample of 41 Canadian adults is between 72.43 and 77.73 litres?
(d) Using the 68-95-99.7 Rule, approximately 95% of samples of 24 Canadian adults will have mean daily water intakes between and .
(e) If we take a random sample of 15 Canadian adults, there is an 11% chance that their mean daily water intake will be greater than
saffsa
haha im in your class and same name haha same question too
(a)You use the squareroot of 2 since there 2 Canadians
P(X<1.65)=P[Z<(1.65-1.86)/(0.29/√2)
Your answer should be P(Z<-0.98), You would then look at the Standard Normal probabilities table, which should give you, 0.1660.
b) would be the same formula but you would use (1.8-1.86) and your squareroot would be √36 instead. Then again you would use the Standard Normal probabilities table, once you get your answer from there you have to do 1-n
n=the answer from Standard Normal probabilities table
c)Step 1: P(72.43/41<X< 77.73/41) P=(1.7666 <X< 1.8959)
Step 2: P[(1.7666-1.86)/(0.29/√41)] and (1.8959-1.86)/(0.29/√41)
Step 3:Once you get your answers again use the Standard Normal probabilities table.
Step 4: Bigger number minus smaller number. That should give you your answer.
d) For the first part
μ= 1.86
σ=0.29
n= 24
Use this formula; μ-2(σ/√n)
For the second part use: μ+2(σ/√n)
(e) Not a 110% about this one sorry. But the formula should be similar as previous, where z=1.23...I think.
x= μ+z(σ/√n)
(a) To calculate the probability that the mean daily water intake for a random sample of two Canadian adults is less than 1.65 litres, we can use the properties of the normal distribution.
First, we need to calculate the standard error of the mean (SEM) for a sample size of 2. The formula for SEM is the standard deviation divided by the square root of the sample size:
SEM = standard deviation / √sample size
In this case, the standard deviation is 0.29 litres and the sample size is 2:
SEM = 0.29 / √2 = 0.205 litres
Next, we can use a Z-table (or software) to find the probability associated with a Z-score of -0.73. The Z-score represents the number of standard deviations below the mean. To calculate the Z-score, we use the formula:
Z = (X - mean) / SEM
In this case, X is 1.65 litres, the mean is 1.86 litres, and SEM is 0.205 litres:
Z = (1.65 - 1.86) / 0.205 = -0.73
Using the Z-table, we find that the probability associated with a Z-score of -0.73 is approximately 0.2314. Therefore, the probability that the mean daily water intake for a random sample of two Canadian adults is less than 1.65 litres is approximately 0.2314.
(b) To calculate the probability that the mean daily water intake for a random sample of 31 Canadian adults is greater than 1.8 litres, we can again use the properties of the normal distribution.
First, we calculate the SEM for a sample size of 31:
SEM = 0.29 / √31 ≈ 0.052 litres
Next, we calculate the Z-score for a mean of 1.8 litres:
Z = (1.8 - 1.86) / 0.052 ≈ -1.15
Using the Z-table, we find that the probability associated with a Z-score of -1.15 is approximately 0.1251. Therefore, the probability that the mean daily water intake for a random sample of 31 Canadian adults is greater than 1.8 litres is approximately 0.1251.
(c) To calculate the probability that the total daily water intake for a random sample of 41 Canadian adults is between 72.43 and 77.73 litres, we need to use the properties of the normal distribution for the sum of random variables.
For the total daily water intake, we need to calculate the mean and standard deviation of the sum, which are obtained by multiplying the mean and standard deviation of the individual variables by the sample size:
Mean_total = mean * sample size = 1.86 * 41 = 76.26 litres
StandardDeviation_total = standard deviation * √sample size = 0.29 * √41 ≈ 1.433 litres
Next, we can calculate the Z-scores for the upper and lower bounds of the range:
Z_lower = (72.43 - 76.26) / 1.433 ≈ -2.67
Z_upper = (77.73 - 76.26) / 1.433 ≈ 1.03
Using the Z-table, we find that the probability associated with a Z-score of -2.67 is approximately 0.0038, and the probability associated with a Z-score of 1.03 is approximately 0.8498. Therefore, the probability that the total daily water intake for a random sample of 41 Canadian adults is between 72.43 and 77.73 litres is approximately 0.8498 - 0.0038 ≈ 0.846.
(d) The 68-95-99.7 rule states that approximately 68% of the data lies within one standard deviation of the mean, 95% lies within two standard deviations, and 99.7% lies within three standard deviations.
To calculate the range within which approximately 95% of samples of 24 Canadian adults will have mean daily water intakes, we can use the formula:
Range = mean ± (2 * SEM)
For a sample size of 24, we calculate the SEM using the given standard deviation:
SEM = 0.29 / √24 ≈ 0.059 litres
The mean is still 1.86 litres.
The lower bound of the range is:
LowerBound = 1.86 - (2 * 0.059) ≈ 1.74 litres
The upper bound of the range is:
UpperBound = 1.86 + (2 * 0.059) ≈ 1.98 litres
Therefore, approximately 95% of samples of 24 Canadian adults will have mean daily water intakes between 1.74 and 1.98 litres.
(e) The question asks for the value above which there is an 11% chance that the mean daily water intake for a random sample of 15 Canadian adults will be greater.
To find this value, we need to use the Z-table and find the Z-score associated with a cumulative probability of 0.11. This Z-score represents the number of standard deviations above the mean.
Using the Z-table, we find that the Z-score associated with a cumulative probability of 0.11 is approximately -1.23.
We can then calculate the mean daily water intake for a sample size of 15:
Mean = 1.86 litres
Next, we calculate the value above which there is an 11% chance of the mean daily water intake being greater:
Value = mean + (Z * SEM)
For a sample size of 15, we can calculate the SEM:
SEM = 0.29 / √15 ≈ 0.075 litres
Using the calculated values:
Value = 1.86 + (-1.23 * 0.075) ≈ 1.749
Therefore, there is an 11% chance that the mean daily water intake for a random sample of 15 Canadian adults will be greater than 1.749 litres.