Sue either travels by bus or walks when she visits the shops. The probability that she catches the bus from the shops is 0.7. Show that the probability that she walks at least one way is 0.72

Corrected Question

Sue travels by bus or walks when she visits the shops. The probability that she catches the bus to the shops is 0.4. The probability she catches the bus from the shops is 0.7. Show the probability that Sue walks at one way is 0.72

(Proved)

Step-by-step explanation:

Sue travels by bus or walks when she visits the shops.

Let the event that she catches the bus to the shop=A

Let the event that she catches the bus from the shop=B

P(A)=0.4

P(B)=0.7

Both A and B are independent events.

Therefore,Probability that she catches the bus to and from the shop:

P(A∩B) = 0.4 X 0.7= 0.28

Probability Sue walks at least one way

Hence, the probability that Sue walks at least one way is 0.72.

c. (3,2,6) is the answer.

d. 24+-0.114

step-by-step explanation:

the question requires us to calculate the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample given the following statistics;

sample mean = 24

standard deviation, σ = 0.4

sample size , n = 49

the confidence interval for a population mean is calculated using the formula;

sample mean ± z-score*(σ/sqrt(n))

the z-score associated with a 95% confidence interval, from the standard normal tables, is;

1.96

substituting these values into the formula;

24 ± 1.96*(0.4/7)

= 24 ± 0.112

from the alternatives given, choice d is closest to the above expression;