This question is incomplete, the complete question is;
Trevor is interested in purchasing the local hardware/electronic goods store in a small town in South Ohio. After examining accounting records for the past several years, he found that the store has been grossing over $850 per day about 60% of the business days it is open. Estimate the probability that the store will gross over $850 at least 6 out of 10 business day
Answer:
P( at least 6 out of 10 business days ) = 0.6312
Step-by-step explanation:
Given the data in the question;
p = 60% = 0.6
q = 1 - p = 1 - 0.6 = 0.4
Using Binomial distribution;
P( X = r ) = [tex]^nC_r p^rq^{n-r[/tex]
where [tex]^nC_r = n!/(r!(n-r))[/tex]
we substitute
P( at leas 6 ) = P(6) + P(7) + P(8) + P(9) + P(10)
= [¹⁰C₆ × 0.6⁶ × 0.4⁴] + [¹⁰C₇ × 0.6⁷ × 0.4³] + [¹⁰C₈ × 0.6⁸ × 0.4²] + [¹⁰C₉ × 0.6⁹ × 0.4¹] + [¹⁰C₁₀ × 0.6¹⁰ × 0.4⁰]
= [0.250] + [0.214] + [0.1209] + [ 0.0403] + [ 0.00604 ]
= 0.6312
Therefore, P( at least 6 out of 10 business days ) = 0.6312
