Suppose the mean income of firms in the industry for a year is 95 million dollars with a standard deviation of 11 million dollars. If income

Question

Suppose the mean income of firms in the industry for a year is 95 million dollars with a standard deviation of 11 million dollars. If incomes for the industry are distributed normally, what is the probability that a randomly selected firm will earn less than 114 million dollars

in progress 0
Gerda 3 years 2021-07-28T11:59:14+00:00 1 Answers 27 views 0

Answers ( )

  1. thanhcong
    0
    2021-07-28T12:00:25+00:00
    Reply

    Answer:

    95.73%

    Step-by-step explanation:

    Given data:

    mean μ= 95

    standard deviation, σ = 11

    to calculate, the probability that a randomly selected firm will earn less than 114 million dollars;

    Use normal distribution formula

    P(X<114)=P(Z<\frac{X-\mu}{\sigma} )

    Substitute the required values in the above equation;

    P(X<114)=P(Z<\frac{114-95}{11} )\\P(X<114)=P(Z<1.7272)\\P(X<114)=0.9573

    Therefore, the probability that a randomly selected firm will earn less than 114 million dollars = 95.73%

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )