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3 July, 08:06

Current research indicates that the distribution of the life expectancies of a certain protozoan is normal with a mean of 48 days and a standard deviation of 10.5 days. Find the probability that a simple random sample of 36 protozoa will have a mean life expectancy of 51 or more days.

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  1. 3 July, 08:24
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    P (x_bar ≥ 51) = 0.0432

    Step-by-step explanation:

    Solution:-

    - The random variable "X" denotes:

    X : life expectancies of a certain protozoan

    - The variable "X" follows normal distribution.

    X ~ Norm (48, 10.5^2)

    - A sample of n = 36 days was taken.

    - The sample is also modeled to be normally distributed:

    x ~ Norm (48, (10.5 / √n) ^2)

    - The sample standard deviation s = 10.5 / √n = 10.5 / √36

    s = 1.75

    - We are to investigate the the probability of sample mean x_bar ≥ 51 days:

    P (x_bar ≥ 51)

    - Standardize the results, evaluate Z-score:

    P (Z ≥ (x_bar - u) / s) = P (Z ≥ (51 - 48) / 1.75)

    P (Z ≥ 1.7142).

    - Use the standardized normal table and evaluate:

    P (Z ≥ 1.7142) = 0.0432

    Hence, P (x_bar ≥ 51) = 0.0432
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