Among the 10 most popular sports, men include competition-type sports – pool and billiards, basketball, and softball – whereas women include

Question

Among the 10 most popular sports, men include competition-type sports – pool and billiards, basketball, and softball – whereas women include aerobics, running, hiking, and calisthenics. However, the top recreational activity for men was still the relaxing sport of fishing, with 41% of those surveyed indicating that they had fished during the year. Suppose 180 randomly selected men are asked whether they had fished in the past year. Suppose 180 randomly selected men are asked whether they has fished in the past year.
a. What is the probability that fewer than 50 had fished?
b. What is the probability that between 50 and 75 (inclusive) had fished?
c. If the 180 men selected for the interview were selected by the marketing department of a sporting-goods company based on information obtained from their mailing lists, what would you conclude about the reliability of their survey results?

in progress 0
Ella 2 weeks 2021-09-10T19:10:55+00:00 1 Answer 0

Answers ( )

    0
    2021-09-10T19:12:50+00:00

    Answer:

    (a) The probability that fewer than 50 had fished is 0.0002.

    (b) The probability that between 50 and 75 (inclusive) had fished is 0.6026.

    (c) The survey results are not reliable.

    Step-by-step explanation:

    Let X = number of men who had fished during the year.

    The probability of the random variable X is, p = 0.41.

    A random sample of n = 180 men are selected.

    The random variable X follows a Binomial distribution with parameters n = 180 and p = 0.41.

    A Normal approximation to Binomial is applied when the following conditions are met,

    1. np ≥ 10
    2. n(1 – p) ≥ 10

    Check:

    np=180\times0.41=73.8>10\\n(1-p)=180\times (1-0.41)=63.72>10

    Thus, the distribution of x can be approximated by a Normal distribution with:

    Mean = np=180\times0.41=73.8

    Standard deviation = \sqrt{np(1-p)}=\sqrt{180\times0.41\times(1-0.41)}=6.599

    (a)

    Compute the probability that fewer than 50 had fished as follows:

    P(X<50)=P(\frac{X-\mu}{\sigma}>\frac{50-73.8}{6.599})\\=P(Z<-3.61)\\=1-P(Z<3.61)\\=1-0.9998\\=0.0002

    Thus, the probability that fewer than 50 had fished is 0.0002.

    (b)

    Compute the probability that between 50 and 75 (inclusive) had fished as follows:

    P(50\leq X\leq 75)=P(49.5<X<75.5)\\=P(\frac{49.5-73.8}{6.599}<\frac{X-\mu}{\sigma}<\frac{75.5-73.8}{6.599})\\=P(-3.68<Z<0.26)\\=P(Z<0.26)-P(Z<-3.68)\\=0.6026-0\\=0.6026

    Thus, the probability that between 50 and 75 (inclusive) had fished is 0.6026.

    (c)

    If the sample of 810 men are selected from mailing list then it is highly probable that the sample is not a representative of the true population, i.e. sports men.

    Because the people interested in sports are less likely to be interested in fishing.

    Thus, it could be concluded that the survey results are not reliable.

Leave an answer

45:7+7-4:2-5:5*4+35:2 =? ( )