Advertisements
Advertisements
प्रश्न
A car hiring firm has two cars. The demand for cars on each day is distributed as a Poison variate, with mean 1.5. Calculate the proportion of days on which neither car is used
Advertisements
उत्तर
In a posion distribution n=2
Mean λ = 1.5
x follows poison distribution
With in P(x) = `("e"^(-lambda) lambda^x)/(x!)`
P(neither car is used) = P(X = 0)
`("e"^(-1.5) (1.5)^0)/(0!) = "e"^(-1.5)`
= 0.2231
APPEARS IN
संबंधित प्रश्न
Define Bernoulli trials
Forty percent of business travellers carry a laptop. In a sample of 15 business travelers, what is the probability that 3 will have a laptop?
Forty percent of business travellers carry a laptop. In a sample of 15 business travelers, what is the probability that 12 of the travelers will not have a laptop?
A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability of 2 successes
Define Poisson distribution
In a test on 2,000 electric bulbs, it was found that bulbs of a particular make, was normally distributed with an average life of 2,040 hours and standard deviation of 60 hours. Estimate the number of bulbs likely to burn for more than 2,150 hours
Choose the correct alternative:
Which of the following statements is/are true regarding the normal distribution curve?
Choose the correct alternative:
In a large statistics class, the heights of the students are normally distributed with a mean of 172 cm and a variance of 25 cm. What proportion of students is between 165cm and 181 cm in height?
Choose the correct alternative:
Cape town is estimated to have 21% of homes whose owners subscribe to the satellite service, DSTV. If a random sample of your home is taken, what is the probability that all four homes subscribe to DSTV?
The birth weight of babies is Normally distributed with mean 3,500g and standard deviation 500g. What is the probability that a baby is born that weighs less than 3,100g?
