After examining daily receipts over the past year, it was found that the green parrot italian restaurant has been grossing over
1 answer:
Complete question is;
After examining daily receipts over the past year, it was found that the green parrot italian restaurant has been grossing over $2200 a day for about 85% of it's business days. Using this as a reasonably accurate measure, find the probability that the green parrot will gross over $2200:
A. At least 5 of the next 7 business days.
B. at least 5 of the next 10 business days.
C. less than 3 of the next 5 business days.
d. Exactly 7 of the next 10 business days
e. At least 1 of the next 5 business days.
Answer:
A) P(X ≥ 5) = 0.9262
B) P(X ≥ 5) = 0.9986
C) P(X ≤ 2) = 0.0266
D) P(X = 7) = 0.1298
E) P(X ≥ 1) = 0.9999
Step-by-step explanation:
This is a binomial probability distribution problem and as such we will use the formula;
P(X = k) = C(n, k) × p^(k) × (1 - p)^(n - k)
A. At least 5 of the next 7 business days will be;
P (X ≥ 5) = P(X = 5) + P(X = 6) + P(X = 7)
P(X = 5) = C(7, 5) × 0.85^(5) × (1 - 0.85)^(7 - 5)
P(X = 5) = 0.20965
P(X = 6) = C(7, 6) × 0.85^(6) × (1 - 0.85)^(7 - 6)
P(X = 6) = 0.396
P(X = 7) = C(7, 7) × 0.85^(7) × (1 - 0.85)^(7 - 7)
P(X = 7) = 0.320577
Thus;
P(X ≥ 5) = 0.20965 + 0.396 + 0.320577
P(X ≥ 5) = 0.9262
B. Probability of at least 5 of the next 10 business days will be;
P(X ≥ 5) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)
From online binomial calculator attached, we can see that the answer is;
P(X ≥ 5) = 0.9986
C. Probability of less than 3 of the next 5 business days will be;
P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)
From online binomial calculator attached, we can see that the answer is;
P(X ≤ 2) = 0.0266
D. Probability of exactly 7 of the next 10 business days will be;
P(X = 7) = C(10, 7) × 0.85^(7) × (1 - 0.85)^(10 - 7)
P(X = 7) = 0.1298
e. Probability of at least 1 of the next 5 business days will be;
P(X ≥ 1) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)
From third image attached using online binomial calculator, we have;
P(X ≥ 1) = 0.9999
You might be interested in
Answer:
a) 0.277 = 27.7% probability that a randomly selected tyre lasts between 55,000 and 65,000 miles.
b) 0.348 = 34.8% probability that a randomly selected tyre lasts less than 48,000 miles.
c) 0.892 = 89.2% probability that a randomly selected tyre lasts at least 41,000 miles.
d) 0.778 = 77.8% probability that a randomly selected tyre has a lifetime that is within 10,000 miles of the mean
Step-by-step explanation:
Problems of normally distributed distributions are solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this question, we have that:
Probabilities:
A) Between 55,000 and 65,000 miles
This is the pvalue of Z when X = 65000 subtracted by the pvalue of Z when X = 55000. So
X = 65000
has a pvalue of 0.954
X = 55000
has a pvalue of 0.677
0.954 - 0.677 = 0.277
0.277 = 27.7% probability that a randomly selected tyre lasts between 55,000 and 65,000 miles.
B) Less than 48,000 miles
This is the pvalue of Z when X = 48000. So
has a pvalue of 0.348
0.348 = 34.8% probability that a randomly selected tyre lasts less than 48,000 miles.
C) At least 41,000 miles
This is 1 subtracted by the pvalue of Z when X = 41,000. So
has a pvalue of 0.108
1 - 0.108 = 0.892
0.892 = 89.2% probability that a randomly selected tyre lasts at least 41,000 miles.
D) A lifetime that is within 10,000 miles of the mean
This is the pvalue of Z when X = 51200 + 10000 = 61200 subtracted by the pvalue of Z when X = 51200 - 10000 = 412000. So
X = 61200
has a pvalue of 0.889
X = 41200
has a pvalue of 0.111
0.889 - 0.111 = 0.778
0.778 = 77.8% probability that a randomly selected tyre has a lifetime that is within 10,000 miles of the mean