Geo-net, a cellular phone company, has collected the following frequency distribution for the length of calls outside its normal customer roaming area: Length (min.) Frequency 0<5 260<5 75 5<10 13910<15 10515<20 3720<25 1825+ 400 The sample mean(x) for this distribution is 14.3 minutes, and the sample standard deviation is 3.7 minutes. Determine whether these data are normally distributed (a = .05).

Reject H0 ; and conclude that call length does not follow a normal distribution.

Step-by-step explanation:

Given :

The hypothesis :

H0: Call lengths outside normal customer roaming areas follows normal distribution

H1: Call lengths outside normal customer roaming areas do not follows normal distribution

Mean, μ = 14.3

Standard deviation, σ = 3.7

From the frequencies Given :

Expected values can be calculated :

Observed values :

16, 75, 139, 105, 37, 18 ; Total = 400

P(Z < (x - μ) / σ)) * total frequency

x = frequency

For x = 5 ;

P(Z < (5 - 14.3) / 3.7)) * 400 = 2.391

For x = 10;

P(Z < (10 - 14.3) / 3.7)) * 400 = 46.644

For x = 15;

P(Z < (15 - 14.3) / 3.7)) * 400 = 180.960

For x = 20;

P(Z < (20 - 14.3) / 3.7)) * 400 = 145.32

For x = 25;

P(Z < (25 - 14.3) / 3.7)) * 400 = 23.92

For x = 30;

P(Z < (30 - 14.3) / 3.7)) * 400 = 0.766

χ² = Σ(O - E)²/E

O = observed values

E = Expected values

χ² = (26-2.391)^2 / 2.391 + (75-46.644)^2 / 46.644 + (139-180.96)^2 / 180.96 + (105-145.32)^2 / 145.32 + (37-23.92)^2 / 23.92 + (18-0.766)^2 / 0.766 = 666.17

χ² = 666.17

The critical value "; df = n - 1= 6-1 = 5

α = 0.05

χ²critical(0.05 ; 5) = 11.07

χ²statistic > χ²critical ; Reject the Null, H0 ; and conclude that call length does not follow a normal distribution.

3

step-by-step explanation: