Question

Statistics and Probability Question. Please show and explain all calculations in detail so that we can learn. Thank you!

Problem 5. Market data from customers who buy Surface Pro 7 or Surface Pro 6 computers show that a randomly selected customer chooses a Surface Pro 7 with a probability of 0.6. Consider 100 randomly selected such customers and assume that each individual independently selects a Surface Pro 7 with a probability of 0.6 and a Surface Pro 6 with a probability of 0.4. a. (3 points) What is the distribution of the number of customers who select Surface Pro 7 among the 100 randomly selected customers? b. (4 points) What are the mean and standard deviation of the number of customers who select Surface Pro 7 among the randomly selected 100 customers? c. (5 points) What is the probability that the number of customers who select Surface Pro 7 is at least one and a half standard deviations above its mean. d. (6 points) A computer store currently has 30 Surface Pro 6 and 80 Surface Pro 7 computers. If the 100 customers come in one by one independently to purchase their favorite computers, what is the probability that all 100 customers get the types of computer they want from the current stock. Show details.

Answer 1

a) since here the buying of the surface pro is treated as a success with a probability (p) of 0.6 and buying surface pro 6 as a failure (q) with probability 0.4 and p=1-q

so here it follows a binomial distribution, i.e, $\binom{n}{x}p^xq^{n-x}$ , here n=100, p=0.6, q=0.4

b)as we know the mean of a binomial distribution as
пр
and standard deviation as $\sqrt npq$

therefore, mean =100*0.6=60

standard deviation= $\sqrt (100*0.6*0.4)$ = 4.89

c) here we have to find

100 PCX 2 mean+0.55d) = P(x > 63) = 2 (9) pan- i=63
$P(X\geq mean+0.5sd)=P(X\geq 63)=\sum_{i=63}^{100}\binom{n}{i}p^iq^{n-i}=0.239$

d)here we have to calculate the probability of 70 and above independent customers getting what they want

as we can see that the minimum no. of smart pro 6 is 30 so to satisfy the 100 customers we calculate the probability as follows

$P(X\geq 70)=\sum_{i=63}^{100}\binom{n}{i}p^iq^{n-i}=0.14775$