Help me solve this question please

Question

The results of a survey by a large supermarket show that 35 % of its customers shop online.

(a) Six customers are chosen at random. Find the probability that more than three of them shop online.

(b) For a random sample of n customers, the probability that at least one of them shops online is greater than 0.95 . Find the least possible value of n.

(c) For a random sample of 100 customers, use a suitable

approximating distribution to find the probability that more than 39 customers shop online.

Answer 1

(a) if p = 0.35 and q = 1-p then this is C(6.4) p⁴ q ² + C(6,5) p⁵ q¹ + C(6,6) p⁶ q ⁰

p (X <= 3) on a TI-84. BINOMCDF. trials = 6, p = 0.35, x value = 3, giving 0.8826

p (X>3) = 1 - 0.8826 = 0.1174 which should agree with the above.

(b) Probability all shop offline = 0.65ⁿ

Probability that at least one shops online = 1 - 0.65ⁿ > 0.95

0.65ⁿ < 0.05

0.65ⁿ = 0.05

n log 0.65 = log 0.05

n = (log 0.05 / log 0.65) = 6.95

^{So n = 7}

Approximate with a normal distribution, mean = np, standard deviation = ^√npq

^{n = 100, p = 0.35, q = 0.65}

p (X > 39)

normalcdf

^{lower = 39}

^{upper = 1E99}

^{mean = 100 x 0.35}

^0.2

^