Given: A college-entrance exam is designed so that scores are normally distributed with a mean $(\mu)$ = 500 and a standard deviation $(\sigma)$ = 100.

A z-score measures how many standard deviations a given measurement deviates from the mean.

Let Y be a random variable that denotes the scores in the exam.

Formula for z-score = $\dfrac{Y-\mu}{\sigma}$

Z-score = $\dfrac{625-500}{100}$

⇒ Z-score = $\dfrac{125}{100}$

⇒Z-score =1.25

Therefore , the required z-score = 1.25

6 0

3 years ago

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 probabi

Ksenya-84 [330]

Answer:

Proved,

P(A∪B)=0.72

Step-by-step explanation:

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

Probability( catch the bus to the shop ), P(A) = 0.4

Probability( catch the bus from the shop ), P(B) = 0.7

Both A and B are independent events.

Therefore,

P(A∩B) = 0.4×0.7

= 0.28

Probability Sue walks one way = 1 - P(A∩B)

= 1 - 0.28

= 0.72

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

3 0

3 years ago

400

Feliz [49]

Answer:

Step-by-step explanation:

7 0

3 years ago